Question

A car travelling at 22 m/s on a straight horizontal road crashes into a tree and comes to a stop. As it does so, the front of the car crumples and the driver moves a distance of 1.07m between impact and rest. What was the acceleration of the driver during this motion? Acceleration is constant

Answer 1

The deceleration of the driver when the car hit the tree and came to a stop was approximately -226.17 m/s².

Step-by-step explanation:

The question asks for the acceleration of a driver when a car crashes into a tree and comes to a stop. Since the initial velocity (vi) of the car is 22 m/s, the final velocity (vf) is 0 m/s (because the car comes to a stop), and the distance (d) over which the deceleration occurred is 1.07 m, we can use the kinematic equation:

vf2 = vi2 + 2ad

We can solve for acceleration (a) as follows:

0 = (22 m/s)2 + 2a(1.07 m)

0 = 484 m2/s2 + 2.14a

-484 m2/s2 = 2.14a

a = -484 m2/s2 / 2.14

a = -226.17 m/s2

The negative sign indicates that this is a deceleration, which is expected since the car is coming to a stop. Thus, the deceleration of the driver was approximately -226.17 m/s2.