A car traveling 56.0 km/h is 24.0 m from 1


A car traveling 56.0 km/h is 24.0 m from a barrier when the driver slams on the brakes.The car hits the barrier 2.00 s later. (a) What is the magnitude of the car’s constant acceleration before impact? (b) How fast is the car traveling at impact?

Solution:

In this question, the car will have a constant negative deceleration in order to avoid hitting the barrier. Let us first convert 56.0 km/h to m/s, 56.0 km/h= 15.55m/s. Now, we can use the following formula:

x-x_0=v_0t_\frac{1}{2}at^2 where x is final displacement, x_0 is initial displacement, a is acceleration, v_0 is initial velocity, and t is time. In our case, the initial displacement is 0, our initial velocity is 15.55 m/s and our time, t=2.00 s. Substituting these values and isolating for a gives us:

a=\frac{2(x-v_{0}t)}{t^2}

a=\frac{2[(24.0m)-(15.55m/s)(2.00s)]}{(2.00s)^2}

a=-3.56m/s^2

As noted previously, the acceleration is negative because the car is slowing down.

 

b) To figure out the speed of the car during impact, we can use the following equation:

v=v_0+at where v is velocity final, v_0 is velocity initial, a is acceleration and t is time. Substituting our values gives us:

v=15.55m/s+(-3.56m/s^2)(2.00s)

v=8.43m/s which is 30.3 km/h.

This question can be found in Fundamentals of Physics, 10th edition, chapter 2, question 33.

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