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A car is traveling on a straight road at a constant 35 m/s, which is faster than the speed limit. Just as the car passes a police motorcycle that is stopped at the side of the road, the motorcycle accelerates forward in pursuit. The motorcycle passes the car 14.5 s after starting from rest.

a). What is the acceleration of the motorcycle (assumed to be constant)?

b). How far does the motorcycle travel before it passes the car?

User Shaniquia
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Answer:

a) 4.8 m/s² b) 505 m.

Step-by-step explanation:

a) As all the movement happens along a straight line, we need to define an axis only, which we call x-axis, being the positive direction the one followed by the car.

We can choose to place our origin at the location where the motorcycle was stopped at the side of the road (assuming that it is the same point for the car when it passes him), so our initial position is 0.

We can also choose our time origin to be the same as the instant that the motorcycle starts from rest, so t₀ = 0.

With these assumptions, and assuming also that the acceleration is constant, we can write two equations, one for the car (at constant speed) and the another one for the motorcycle, as follows:

xc = vx*t

xm= 1/2*a*t²

When the motorcycle passes the car, both distances traveled from the origin will be equal each other, i.e., xc = xm :

⇒ vx*t = 1/2*a*t²

We have as givens vx=35 m/s and t = 14.5 sec when both equations are equal each other.

⇒ 35 m/s* 14.5 s = 1/2*a*(14.5)²(s)²

Solving for a:

a = (2* 35 m/s) / 14.5 s = 4.8 m/s²

b) Replacing the value of a in the equation for xm, we have:

xm = 1/2*4.8 m/s²* (14.5)²s² = 505 m.

User Yanay
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