Question

question 10
a car is moving at a constant velocity when it is involved in a collision. the car comes to rest after 0.450 s with an average acceleration of 60.0 m/s² in the direction opposite that of the cars velocity. what was the speed, in km/h, of the car before the collision?
105 km/h
44.8 km/h
97.2 km/h
144 km/h
29.2 km/h

Explanation:

Step1: Use the kinematic - equation

We use the equation $v = v_0+at$. Since the car comes to rest, $v = 0$. The acceleration $a=- 60.0\ m/s^{2}$ (negative because it's in the opposite direction of the initial velocity) and $t = 0.450\ s$. We solve for $v_0$.
$0=v_0+(-60.0\ m/s^{2})\times0.450\ s$
$v_0 = 60.0\ m/s^{2}\times0.450\ s=27\ m/s$

Step2: Convert the speed to km/h

We know that $1\ m/s=3.6\ km/h$.
So, $v_0=27\ m/s\times3.6\ km/h\div1\ m/s = 97.2\ km/h$

Answer:

C. 97.2 km/h