Question

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.51 m/s2 for t1 = 20 s. At that point the driver notices a tree limb that has fallen on the road and brakes hard for t2 = 5 s with a constant acceleration of a2 = -5.03 m/s2. Write an expression for the car's speed just before braking.

Answer 1

v = 1.51t and v = 30.2 m/s just before braking

Step-by-step explanation:

Using v = u + at where u = initial velocity of car = 0 m/s (since it starts from rest), a = a1 = acceleration before braking = 1.51 m/s², t = t1 = time before braking = 20s and v = speed of car before braking.

Substituting the values of the variables into the equation, we have

v = u + at

v = 0 m/s + 1.51 m/s²t

v = 1.51t and its value after 20 s is

v = 1.51 × 20

v = 30.2 m/s which is the value just before braking.