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Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Physics 1101: Kinematics Solutions

  1. Initially, a ball has a speed of 5.0 m/s as it rolls up an incline. Some time later, at a distance of 5.5 m up the incline, the ball has a speed of 1.5 m/s DOWN the incline.
    (a) What is the acceleration? What is the average velocity? How much time did this take?
    (b) At some point the velocity of the ball had to have been zero. Where and when did this occur?

    A well-labeled sketch usually helps make the problem clearer.

    (a) Next, we list the list the given information and what we are looking for:

    v0 = +5.0 m/s
    vf = −1.5 m/s
    Δx = +5.5 m
    a = ?
    vaverage = ?
    t = ?

    Note that I have taken the direction up the incline as positive and that the signs are explicitly stated. It is a very common source of error to leave out or to not consider the signs of directions of all vector quantities.

    To find the acceleration, we find the kinematics equation that contains a and the given quantities. Examining our equations we see that we can use . Rearranging this equation to find a yields . Notice that the acceleration is negative. This means that the acceleration points down the incline. It means that an object traveling up an incline will slow, turn around, and roll down the incline.

    The average velocity is defined .

    To find the time, we find the kinematics equation that contains a and the given quantities. Examining our equations we see that we can use . Rearranging this equation to find t yields .

    (b) When an object moving in 1D turns around we know that the object is instantaneously at rest and that its velocity at that point is v3 = 0. The information that we know is thus:

    v0 = +5.0 m/s
    v3 = 0 m/s This is our new final velocity
    a = −2.068 m/s2       From part (a)
    Δx = ?
    vaverage = ?
    t = ?

    Notice that the acceleration is a constant of the motion; it has the same value in both parts of the problem.

    To find the displacement from the initial position where the ball turns around, we find the kinematics equation that contains x and the given quantities. Examining our equations we see that we can use . Rearranging this equation to find x yields . Notice that this value is bigger than the original 5.5 m and is consistent with the sketch, i.e. the ball was farther up the incline when it turned around.

    To find the time it takes for the ball to reach the point where it turns around, we find the kinematics equation that contains t and the given quantities. Examining our equations we see that we can use . Rearranging this equation to find t yields .Notice that this value is smaller than the time in part (a) and is consistent with the sketch, i.e. the ball hasn't come back down the incline yet.

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  2. A bullet in a rifle accelerates uniformly from rest at a = 70000 m/s2. If the velocity of the bullet as it leaves the muzzle is 500 m/s, how long is the rifle barrel? How long did it take for the bullet to travel the length of the barrel? What is the average velocity of the bullet?

    To solve this problem, we list the given information and what we are looking for:

    v0 = 0.0 m/s since the bullet is initially at rest
    vf = 500 m/s velocity of the bullet as it leaves the barrel
    a = 70,000 m/s2      
    Δx = ? the length of the barrel
    t = ? the time it takes to travel the barrel
    vaverage = ?

    To find the length of the barrel, we find the kinematics equation that contains x and the given quantities. Examining our equations we see that we can use . Rearranging this equation to find a yields .

    To find the time it takes for the bullet to travel the length of barrel, we find the kinematics equation that contains t and the given quantities. Examining our equations we see that we can use . Rearranging this equation to find t yields .

    The average velocity is defined .

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  3. A red car is stopped at a red light. As the light turns green, it accelerates forward at 2.00 m/s2. At the exact same instant, a blue car passes by traveling at 62.0 km/h. When and how far down the road will the cars again meet? Sketch the d versus t motion for each car on the same graph. What was the average velocity of the red car for this time interval? For the blue car? Compare the two and explain the result?

    To solve this problem, we list the list the given information

    Red Car Blue Car
    v0 red = 0.0 m/s v0 blue = 62.0 km/h = 17.222 m/s
    ared = 2.00 m/s2       ablue = 0 m/s2 (constant velocity)
    Δxred = ? Δxblue = ?
    tred = ? tblue = ?

    This is an example of a two-body constrained kinematics problem. Even if a sketch was not explicitly required, we would need one anyway to get the constraints. For the sketch, recall that on a d versus t curve an object moving forward with a uniform acceleration should be represented by a line curving upwards while an object with constant forward velocity is represented by a straight line with a positive slope.


    Looking at the sketch, we see that our constraints are:

    Δxred = Δxblue (1), and

    tred = tblue (2). .

    To solve the problem, we must find the kinematics equation that contains the known quantities, v0 and a, and the unknown quantities, Δx and t. Examining our equations we see that we can use Δx= v0t + ½at2. We substitute this equation into both sides of equation (1). This yields,

    v0 redtred + ½ared(tred)2 = v0 bluetblue + ½ablue(tblue)2.

    We then use equation (2) to replace tred and tblue by t,

    v0 redt + ½aredt2 = v0 bluet + ½abluet2.

    Plugging in the values of the given quantities yields,

    ½(2.00)t2= 17.2t.

    The solution of this equation is t = 17.222 seconds. This is the time that elapses before the two cars meet again.

    With a value for t, we can find how far down the road the red car has traveled;

    Δxred = v0 redt + ½aredt2 = ½(2.00)(17.2)2 = 297 m.

    As a check, we can find how far down the road the blue car has traveled;

    Δxblue = v0 bluet + ½abluet2 = (17.2)(17.2) = 297 m.

    So the cars meet 297 m down the road.

    According to our definition of average velocity, vaverage red= Δxred/t = (297 m)/(17.2 s) = 17.2 m/s. Since the blue car maintains a constant velocity, vaverage blue= v0 blue = 17.2 m/s. The two quantities are the same since the two cars have traveled the same distance in the same amount of time.

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  4. A speeding motorist traveling down a straight highway at 110 km/h passes a parked patrol car. It takes the police constable 1.0 s to take a radar reading and to start up his car. The police vehicle accelerates from rest at 2.1 m/s2. When the constable catches up with the speeder, how far down the road are they and how much time has elapsed since the two cars passed one another?

    To solve this problem, we list the list the given information

    Constable Motorist
    v0 police = 0.0 m/s v0 speeder = 110 km/h = 30.556 m/s
    apolice = 2.00 m/s2       aspeeder = 0 m/s2 (constant velocity)
    Δxpolice = ? Δxspeeder = ?
    tpolice = ? tspeeder = ?

    This is an example of a two-body constrained kinematics problem. We need a sketch to get the constraints. For the sketch, recall that on a d versus t curve an object moving forward with a uniform acceleration should be represented by a line curving upwards while an object with constant forward velocity is represented by a straight line with a positive slope.


    Looking at the sketch, we see that our constraints are:

    Δxspeeder = Δxpolice (1), and

    tspeeder = tpolice + 1 (2). .

    To solve the problem, we must find the kinematics equation that contains the known quantities, v0 and a, and the unknown quantities, Δx and t. Examining our equations we see that we can use Δx= v0t + ½at2. We substitute this equation into both sides of equation (1). This yields,

    v0 speedertspeeder + ½aspeeder(tspeeder)2= v0 policetpolice + ½apolice (tpolice)2.

    We then use equation (2) to replace tspeeder by tpolice + 1,

    v0 speeder (tpolice + 1) + ½aspeeder(tpolice + 1)2 = v0 policetpolice + ½apolice (tpolice)2.

    Plugging in the values of the given quantities yields,

    (30.556)( tpolice + 1) = ½(2.1)(tpolice)2.

    This is a quadratic in tpolice. Solving the quadratic yields, tpolice = 30.07 seconds. It takes the police constable 30.1s to catch up with the speeder. The speeder was traveling for 31.1 s.

    With a value for tpolice, we can find how far down the road the police car has traveled;

    Δxpolice = v0 policetpolice + ½apolice (tpolice)2 == ½(2.1)(30.07)2 = 949 m.

    As a check, we can find how far down the road the speeder's car has traveled;

    Δxspeeder = v0 speeder (tpolice + 1) + ½aspeeder(tpolice + 1)2 = 30.55631.07 = 949 m.

    So the cars meet 949 m down the road.

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  5. A ball is thrown up into the air with an initial velocity of 12.0 m/s. How long will it be in air until it returns to the hand? To what maximum height will it rise?

    To solve this problem, we list the given information and what we are looking for:

    v0 = 12.0 m/s velocity as it leaves the hand
    vtop = 0 m/s since it turns around
    vf = -12.0 m/s symmetry says it must have this value when it returns to the same height
    a = -9.81 m/s only gravity is acting
    Δy = 0 since it returns to the same height
    tair = ? the time it takes for the entire trip
    tup = tdown = ½tair = ?       symmetry requires this


    We have lots and lots of information from symmetry. To find tair, choose the kinematics equation that has t and the known quantities v0, vf, and a, that is vf = v0+ atair. Solving yields tair = (vf - v0)/a = (-v0-v0)/(-g) = 2v0/g = 2.4465 seconds. Hence tup = tdown = 1.2232 s.

    To find h, choose the kinematics equation that has Δy (h is a displacement) and the known quantities v0, vtop, and a, that is . Upon rearrangement, this yields h = Δy = (v0)2/g = 7.34 m.

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  6. A ball is thrown up into the air and returns to the same level. It is in the air for 3.20 seconds. With what initial velocity was it thrown? How high did it rise?

    To solve this problem, we list the list the given information and what we are looking for:

    v0 = ? velocity as it leaves the hand
    vtop = 0 m/s since it turns around
    vf = -v0 symmetry says it must have this value when it returns to the same height
    a = -9.81 m/s only gravity is acting
    Δy = 0 since it returns to the same height
    tair = 3.20 s the time it takes for the entire trip
    tup = tdown = ½tair = 1.60 s       symmetry requires this


    We have lots and lots of information from symmetry. To find v0, choose the kinematics equation that has v0 and the known quantities, vf = -v0 , tair and a, that is vf = v0+ atair. Eliminating vf yields -v0 = v0 - gtair. Rearranging gives v0 = gtair/2 = 15.7 m/s.

    To find h, choose the kinematics equation that has Δy (h is a displacement) and the known quantities v0, vtop, and a, that is . Upon rearrangement, this yields h = Δy = (v0)2/g = 12.6 m.

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  7. Two balls are thrown upwards from the same spot 1.15 seconds apart. The first ball had an initial velocity of 15.0 m/s and the second was 12.0 m/s. At what height do they collide?

    To solve this problem, we list the list the given information

    Ball #1 Ball #2
    v0 1 = 15.0 m/s      v0 2 = 12.0 m/s
    a1 = -9.81 m/s2 a2 = -9.81 m/s2
    Δy1 = ? Δy2 = ?
    t1 = ? t2 = ?

    This is an example of a two-body constrained kinematics problem. We need a sketch to get the constraints. For the sketch, recall the shape of the d versus t curve for an object thrown up into the air - a parabola.


    Looking at the sketch, we see that our constraints are:

    Δy1 = Δy2 (1), and

    t1 = t2 + 1.15 (2). .

    To solve the problem, we must find the kinematics equation that contains the known quantities, v0 and a = -g, and the unknown quantities, Δy and t. Examining our equations we see that we can use Δy = v0t - ½gt2. We substitute this equation into both sides of equation (1). This yields,

    v01t1 - ½g(t1)2 = v02t2 - ½g(t2)2.

    We then use equation (2) to replace t1 by t2 + 1.15,

    v01(t2+1.15) - ½g(t2+1.15)2 = v02t2 - ½g(t2)2.

    This reduces to

    1.15v01 + v01t2 - ½g[(t2)2 +2.30t2 +1.3225] = v02t2 - ½g(t2)2.

    Upon rearrangement this becomes

    (v01 - v02 - 1.15g)t2 = -(1.15v01 -0.66125g) .

    Thus t2 = 1.2997 s, and t1 = 2.4497 s. Now that we have the time that each ball is in the air, we can now find h

    h = v01t1 - ½g(t1)2 = (15.0)(2.4497) - ½g(2.4497)2 = 7.31 m ,

    and double-checking our result

    h = v02t2 - ½g(t2)2 = (12.0)(1.2297) - ½g(1.2297)2 = 7.31 m .

    So the balls collide when they are 7.31 m in the air.

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  8. You are trapped on the top of a burning building. Death is imminent and help is nowhere in sight. There is a safe building 6.50 m away and 3.00 m lower. You decide to try and make it across. You run horizontally off your building at 8.10 m/s. Do you make it across? If you don't, how much faster must you be going?

    9. First we sketch the situation and possible outcomes.

    While you are jumping, you are a projectile. We solve projectile motion problems by considering the x and y components separately, keeping in mind that the time in air is common. We write out the i and j information in separate columns including the information that we can infer or that we are supposed to know like the fact that running horizontally implies that voy = 0..
    i
    j
    Δxsafe = 6.50 m Δysafe = -3.00 m  minus indicates down
    ax = 0 No x component for  projectiles  ay = -g = -9.81 m/s2  gravity acts down
    v0x = 8.10 m/s v0y = 0 m/s horizontal takeoff means
    no vertical component 
    tair = ? common  tair = ?
    Looking at the x information, we see that we have enough data to find tair. The kinematics equation that has all four quantities is Δx = v0xt + ½axt2. Since ax = 0 for a projectile, this equation become Δx = v0xt. Solving for t, we get

    t = Δx / v0x = (6.50 m) / (8.10 m/s) = 0.8025 s .

    This is the time it would take you to cross a horizontal distance of 6.50 m. You must be in the air for at least this long if you are to safely make it across to the next building.

    On the other hand, looking at the y information, we see that we also have enough data to find tair. The kinematics equation that has all four quantities is Δy = v0yt + ½ayt2. We know v0y = 0 since you ran off the roof horizontally and that ay = -g, thus this equation becomes Δy = -½gt2. Solving for t, we get

    t = { -2Δy / g }½ = {(-2×-3.00 m) / (-9.81 m/s) }½ = 0.7821 s .

    This is the time it takes you to fall a vertical distance of 3.00 m. If you do reach the other building, then this is how long you were in the air.

    Since the time it takes to cross the horizontal distance is less than the time you have, you have don't make it across.

    To make it across safely you of course would need to run off the roof faster. Since ax = 0 for projectiles, the kinematics equation become Δx = v0xt where t is now the 0.7821 s. Solving for v0x, we get ,

    v0x = Δx / t = (6.50 m) / (0.7821 s) = 8.31 m/s .

    If you were able to run at 8.31 m/s you would safely make it to the other building.

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  9. A stunt motorcyclist is trying to jump over fifteen buses set side to side. Each bus is 2.50 m wide and a 30.0° ramp has been installed on either side of the line of buses. What is the minimum speed at which she must travel to safely reach the other side. How long will she be in the air?

    10. First we sketch the situation and possible outcomes.

    While the motorcyclist is jumping, she is a projectile. We solve projectile motion problems by considering the x and y components separately, keeping in mind that the time in air is common. We write out the i and j information in separate columns including the information that we can infer or that we are supposed to know. We see from the sketch that if the motorcyclist is successful, then this is an example of level-to-level flight and Δy = 0. Note that the initial velocity is broken into components.
    i
    j
    Δxsafe = 15×2.50 m = 37.5 m  Δysafe = 0  level to level flight
    ax = 0 No x component 
    for projectiles     		 ay = -g = -9.81 m/s2  gravity acts down
    v0x = v0 cosθ v0 is unknown  v0y = v0 sinθ
    tair = ? common  tair = ? common 
    Looking at the x any y information, we see that we have two unknowns, v0 and t, for both. While we cannot solve any equation for x or y since there are two unknowns, both can be solved together. The appropriate kinematics equations that has all four quantities for x and for y is:

    Δx = v0xt + ½axt2 , and Δy = v0yt + ½ayt2.

    We substitute in known quantities to get

    Δx = v0 cosθ t , and 0 = v0 sinθ t - ½gt2.

    We can divide the second equation by t and we get

    Δx = v0 cosθ t, and v0 sinθ = ½gt.

    We rewrite the first equation as t = Δx / v0 cosθ, which we substitute into the second equation to get v0 sin= ½g[Δx / v0 cosθ]. Getting v0 by itself we have v0 = {g/(2sinθ cosθ)}½. Plugging in the appropriate numbers, we get v0 = 20.61 m/s = 74.2 km/h. Since is the speed that the motorcyclist must have on liftoff to successfully reach the other ramp.

    We can substitute this value into t = Δx / v0 cos to find the time in air to be 2.10 s.

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  10. A boy throws a rock with speed v = 18.3 m/s at an angle of θ = 57.0° over a building. The rock lands on the roof 22.0 m in the x direction from the boy. How long was the rock in the air? How much taller, height h, is the building than the boy? Ignore air resistance.

    11. The rock is a projectile. We solve projectile motion problems by considering the x and y components separately, keeping in mind that the time in air is common. We write out the i and j information in separate columns including the information that we can infer or that we are supposed to know. We see from the sketch that h = Δy, the vertical displacement. Note that the initial velocity is broken into components.

    i
    j
    Δx = 22.0 m  Δy = h 
    ax = 0 No x component
    for projectiles     		 ay = -g = -9.81 m/s2  gravity acts down
    v0x = v0cos(θ)
          = 18.3 × cos(57º)
          = 9.9669 m/s     		 v0y = v0sin(θ)
          = 18.3 × sin(57º)
          = 15.3477 m/s 
    tair = ? common  tair = ? common 
    Looking at the x information, we see that we have enough data to find tair. The kinematics equation that has all four quantities is Δx = v0xt + ½axt2. Since ax = 0 for a projectile, this equation become Δx = v0xt. Solving for t, we get

    t = Δx / v0x = (15.0 m) / (9.9669 m/s) = 2.2073 s .

    Looking at the y information, we see that we now have enough data to find h. The kinematics equation that has all four quantities is Δy = v0yt + ½ayt2. Since Δy = h and ay = -g, this equation become h = v0yt - ½gt2. Substituting in the appropriate numbers reveals that h = 9.98 m. The building is 10.0 m taller that the boy where we have assumed that the ball left the boy's hand at head height which is a reasonable assumption.

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  11. A tile, initially at rest, slides down a roof for a distance of 3.75 m before falling off the roof. The height of the building from ground to eave is 8.40 m. The acceleration of the tile as it slides is 2.10 m/s2.

    12. (a) Determine the speed of the tile just as it leaves the roof.
    (b) Determine the vertical component of the velocity just before it leaves the roof.
    (c) Determine the horizontal component of the velocity just before it leaves the roof.
    (d) Determine how long it takes to hit the ground after leaving the roof.
    (e) Determine how far from the edge of the roof that the tile lands.

    When the tile slides down the roof, it travels in a straight line. That is a 1D kinematics problem. When it leaves the roof, it becomes a projectile problem.

    (a) We solve the 1D problem first. We write down all the given data and unknowns:
    Δx = 3.75 m
    v0 = 0 starts from rest 
    a = 2.10 m/s2
    vf need this for the second part 
    Inspecting the data, we see that we can use the kinematics equation 2aΔx = (vf)2 - (v0)2. Solving for vf, we find

    vf = {2aΔx + (v0)2 }½ = {2(3.75 m/s2)(3.75 m) + (0)2 }½ = 3.9686 m/s .

    This is the speed that the tile leaves the roof and is the initial velocity for the second part of the problem.

    (b) The vertical component of the velocity of the tile as it leaves the roof is v0x = -vfsin(25°) = -1.6772 m/s. Note that the minus sign indicates that the tile is moving downwards.

    (c) The horizontal component of the velocity of the tile as it leaves the roof is v0x = vfcos(25°) = 3.5968 m/s.

    (d) The tile is now a projectile. We solve projectile motion problems by considering the x and y components separately, keeping in mind that the time in air is common. We write out the i and j information in separate columns including the information that we can infer or that we are supposed to know. We see from the sketch that Δy is the vertical distance that the tile falls. Note that the initial velocity is broken into components.
    i
    j
    Δx = ?  Δy = -8.40 m minus means down
    ax = 0 No x component 
    for projectiles     		 ay = -g = -9.81 m/s2  gravity acts down
    v0x = 3.5968 m/s v0y = -1.6772 m/s
    tair = ? common  tair = ? common 
    Looking at the y information, we see that we have enough data to find tair. The kinematics equation that has all four quantities is Δy = v0yt +½ayt2. Substituting in the appropriate numbers reveals that we have a quadratic in t:

    -8.40 = -1.6772t + ½(-9.81)t2 .

    The two solution to this quadratic are t = 1.149 s and t = -1.491 s. We take the positive solution as that is the solution for times after the tile left the roof.

    (e) Looking at the x information, we see that we now have enough data to find Δx. The kinematics equation that has all four quantities is Δx = v0xt + ½axt2. Since ax = 0 for a projectile, this equation become Δx = v0xt. Substituting in the appropriate numbers, we get Δx = v0xt = 3.5968 m/s × 1.149 s = 4.13 m. The tile land 4.13 m from the eave of the roof.

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  12. A boy is on the side of a hill. The hill makes a 15° incline with respect to horizontal. The boy throws a rock up the side of the hill. The boy throws the rock at 45° with respect to horizontal and the rock lands 30 m away up the hill. Find how fast the boy threw the rock. What angle does the rock make with horizontal before it lands? Ignore the boy's height.

    13. We know

    v0 = i v0cos(45°) + j v0 sin(45°)
    vf = i vf cos(θ) – j vf sin(θ)
    a = – j g = – j 9.81 m/s2
    Δr = i 30 cos(15°) + j 30 sin(15°) m
         = i 28.978 + j 7.765 m

    Applying our kinematic equations to the situation:

    vf cos(θ) = v0 cos(45°) (1)
    28.978 = v0 cos(45°) t (2)
    7.765 = v0 sin(45°)t – ½gt2 (3)
    –vf sin(θ) = v0 sin(45°) – gt (4)
    2(9.81)(7.765) = [–vf sin(θ)]2 – [v0 sin(45°)]2 (5)

    Examining the equations, we find that we can eliminate v0t from equation (3) using equation (2). This yields

    7.765 = [28.978/cos(45°)] sin(45°) – ½gt2 .

    This can be simplified and solved for t.  We find t = 2.0796 s. Knowing t, equation (2) can be solved to yield v0 = 19.706 m/s. With v0 and t, equations (1) and (4) become vf cos(θ) = 13.934 and vf sin(θ) = 6.467. The ratio of these two equations yields tan(θ) = 0.4641 or θ = 24.9°. Also vf = 15.36 m/s.

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  13. You are 6.0 m from the wall of a house. You want to toss a ball to your friend who is 6.0 m from the opposite wall. The throw and catch each occur 1.0 m above the ground.

    (a) What minimum velocity, expressed in ij notation, will allow the ball to clear the roof?
    (b) At what angle and speed do you throw the ball?


    14. The main point of interest is the rooftop. To just clear the roof, requires that the roof be at the top of the parabola. We know

    v0 = i v0 cos(θ) + j v0 sin(θ)
    vf = i vf + j 0
    a = – j g = – j 9.81 m/s2
    Δr = i 9 + j 5 m

    Applying our kinematic equations to the situation:

    vf = v0 cos(θ) (1)
    9 = v0 cos(θ) t (2)
    5 = v0 sin(θ)t – ½gt2 (3)
    0 = v0 sin(θ) – gt (4)
    2(–9.81)(5) = –[v0 sin(θ)]2 (5)

    Equation (5) allows us to find the y component of the initial velocity v0y = v0 sin(θ) = 9.905 m/s. We can use this result to find t from equation 4, t = 1.0096 s. Knowing t we can use equation (2) to find the x component of the initial velocity v0x = v0 cos(θ) = 8.914 m/s. Thus the initial velocity is

    v0 = i 8.914 + j 9.905 m/s .

    The angle θ can be found from the ratio [v0 sin(θ)] / [v0 cos(θ)] = 9.905/8.914 or, more simply, tan(θ) = 1.111 which yields θ = 48.0°.

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  14. A boy throws a ball horizontally from a shoulder height of 1.10 m. Just before the ball touches down on the level ground it makes an angle of 30° with the ground. Determine the initial velocity of the ball as it left the boy's hand.

    15. We know

    v0 = i v0 + j 0
    vf = i vf cos(30°) – j vf sin(30°)
    a = – j g = – j 9.81 m/s2
    Δr = i Δx – j 1.10 m

    Applying our kinematic equations to the situation:

    v0 = vf cos(30°) (1)
    Δx = v0 t (2)
    –1.10 = –½gt2 (3)
    – vf sin(30°) = –gt (4)
    2(–9.81)(-1.10) = [–vf sin(30°)]2 (5)

    To determine v0, we need to know vf or Δx in equation (1) or (2). We can use equation (5) to find vf = 9.291 m/s. Thus, from equation (1), v0 = 8.046 m/s.

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  15. The earth is 1.50 × 1011 m from the sun. It takes 365 days for the earth to travel in a circle around the sun. What centripetal acceleration is felt by the earth?

    We know that centripetal acceleration is given by the formula ac = v2/R where R is the distance from the earth to the sun which is given and v is the speed of the earth which is not directly given. However, since the earth travels in an almost perfect circle around the sun, we can find the speed from v = d/t = 2πR/T. The time T is one year which we must convert to seconds.

    T = 365 days × 24 hours/day × 3600 seconds/ hour = 31,536,000 s

    The speed is thus

    v = 2πR/T = 2π(1.50 × 1011 m)/(31,536,000 s) = 29,885.8 m/s

    The centripetal acceleration is thus ac = v2/R = (29,885.8 m/s)2/(1.50 × 1011 m) = 0.00595 m/s2. This is tiny compared to g = 9.81 m/s2 and is not usually noticeable.

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