Question

28. (a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32° ramp at a speed of 40.0 m/s (144 km/h) . How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 20.0 m long? (b) Discuss what your answer implies about the margin of error in this act—that is, consider how much greater the range is than the horizontal distance he must travel to miss the end of the last bus. (Neglect air resistance.)

Answer 1

Answer:

a) # buses = 7  

Explanation:

For this exercise we use the kinematic equations, let's find the time it takes to reach the same height

   

     y =$v_{oy}$  t - ½ g t²

Let's decompose the speed, with trigonometry

      v₀ₓ = v₀ cos θ

      $v_{oy}$ = v₀ sin  θ

      v₀ₓ = 40 cos 32

      v₀ₓ = 33.9 m / s

      $v_{oy}$ = 40 sin32

      $v_{oy}$ = 21.2 m / s

When it arrives it is at the same initial height y = 0

         0 = ($v_{oy}$ - ½ gt) t

That has two solutions

       t = 0                    when it comes out

       t = 2 $v_{oy}$ / g       when it arrives

       t = 2 21.2 /9.8

       t = 4,326 s

We use the horizontal displacement equation

       x = vox t

       x = 33.9   4.326

       x = 146.7 m

To find the number of buses we can use a direct proportions rule

    # buses = 146.7 / 20

    # buses = 7.3

    # buses = 7

The distance of the seven buses is

     L = 20 * 7 = 140 m

b) let's look for the scope for this jump

     R = vo2 sin2T / g

     R = 40 2 without 2 32 /9.8

     R = 146.7 m

As we can see the range and distance needed to pass the seven (7) buses is different there is a margin of error of 6.7 m in favor of the jumper (security)