You are driving slowly in the right lane of a straight country road. For a while

Question

You are driving slowly in the right lane of a straight country road. For a while, a car to your left has lagged 53.0 m behind you at the same speed of 22.0 mi/h . Suddenly that car speeds up and passes you, traveling at a constant acceleration until it is 40.0 m in front of you 8.00 s later. You are driving slowly in the right lane of a straight country road. For a while, a car to your left has lagged 53.0 m behind you at the same speed of 22.0 mi/h . Suddenly that car speeds up and passes you, traveling at a constant acceleration until it is 40.0 m in front of you 8.00 s later.

Part A Determine the other car’s acceleration.

Part B How far did each of you travel during the passing procedure? you and other car

Part D What is the other car’s speed at the end of the passing procedure?

Part E) A 2.05-m-tall basketball player takes a shot when he is 6.02 m from the basket (at the three-point line). If the launch angle is 31 and the ball was launched at the level of the player’s head, what must be the release speed of the ball for the player to make the shot? The basket is 3.05 m above the floor.

Explanation / Answer

v = 22 mi/h

= 22*1.609*5/18

= 9.83 m/s

Distance covered by the car on the right

d= v*t

d= 9.83*8

d = 78.64 m---- Distance covered by the car on the right during the passing procedure

Total distance covered by the other car is = 53+78.64+40=171.64 m ---Total distance covered by the car on the left during the passing procedure

Now we use

s = vo*t +0.5 a t^2

171.64 = 9.83*8 +0.5*a*8^2

a=2.91m/s^2------Acceleration of the car

Now the Final Velocity

VF= Vo +a*t

Vf= 9.83 +2.91*8 = 33.11m/s-----Final Velocity

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Second Problem

from the problem the x and y components ofthe initial velocity (v) are

   vix = vicos31o

        = 0.857*vi

   viy = visin31o

        = 0.515*vi

   as the player is 6.02 mfrom the basket the timetaken by the ball to move this distance is

   t = x / vix

     = (6.02 m) /0.9135 vi

   at this time the vertical displacement of theball willl be

   s = (6.02 m - 2.05 m)

      = 3.97 m

   applying the equation of motion we get

   s = viy t + (1 / 2) ayt2 we get    

   ay = - 9.80 m / s2

   solve for vi

   vi = .......... m / s

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