Question

(a) The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of -5.05 m/s² for 4.00 s, making straight skid marks 62.9 m long, all the way to the tree. With what speed (in m/s) does the car then strike the tree? ___________ You have calculated the initial velocity of the car. m/s (b) What If? If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what would the acceleration need to be in m/s²) so that the car narrowly avoids a collision? ____________ m/s²

Answer 1

(a) To find the initial speed at which the car hit the tree, we will use the formula for final velocity, which is given as:

v_final = v_initial + a*t

In this case, the final velocity is zero since the car hits the tree and comes to rest. We can rearrange the equation to solve for 'v_initial' which results in:

v_initial = v_final - a*t

Inserting the given values:

v_initial = 0 - (-5.05 m/s^2 * 4.00 s)

This simplifies to:

v_initial = 20.2 m/s

Therefore, the speed at which the car strikes the tree is 20.2 m/s.

(b) To determine the necessary acceleration for the car to avoid a collision, we'll utilize the second equation of motion which is:

s= ut + (1/2)at^2

Here, u = initial_velocity, a = final_acceleration , s = distance ,and t = time. We can rearrange the equation to solve for 'a', yielding:

a = (2s - 2ut) / t^2

With the known values inserted, the equation becomes:

a = (2*62.9 m - 2*20.2 m/s * 4 s) / (4 s)^2

This simplifies to:

a = -2.2375 m/s^2

Hence, to avoid a collision, the car would need to decelerate at a rate of -2.2375 m/s^2. The negative sign indicates that the acceleration is directed opposite to the motion of the car (deceleration).