Question

a driver speeds along a curved road and sees a fallen tree on the road. he applies the brakes but is unable to stop in time. when the car hits the tree, it slows down from 6 meters/second to 0 meters/second. what was the change in velocity during this time? when the car was stopped by the tree, its change in velocity during the collision was -6 meters/second. this change in velocity occurred in 2 seconds. what was the acceleration of the car? t = s u = m/s v = m/s a =? a = (v - u)/t = -6 m/s / 2 = m/s² velocity change = v - u = 0 - 6 m/s = m/s

Explanation:

Step1: Identify initial and final velocities for the braking - phase

The car slows down from an initial velocity $u = 6$ m/s to a final velocity $v = 0$ m/s.

Step2: Calculate the change in velocity

The formula for the change in velocity $\Delta v=v - u$. Substituting the values, we get $\Delta v=0 - 6=- 6$ m/s.

Step3: Identify values for the collision - phase

During the collision, the change in velocity occurs in $t = 2$ s. The change in velocity $\Delta v=-6$ m/s (already calculated).

Step4: Calculate the acceleration during the collision

The formula for acceleration is $a=\frac{v - u}{t}=\frac{\Delta v}{t}$. Substituting $\Delta v=-6$ m/s and $t = 2$ s, we have $a=\frac{-6}{2}=-3$ m/s².

Answer:

The change in velocity during the braking is -6 m/s. The acceleration during the collision is -3 m/s².