Question

Calculate the acceleration of a 1400-kg car that stops from 31 km/h "on a dime" (on a distance of 1.7 cm).

A. Determine the acceleration of the car.

B. How many g's is the magnitude of the acceleration of the car?

C. What is the force felt by the 67-kg occupant of the car?

Answer 1

Part A.

Using 3rd kinematic equation:

V^2 = U^2 + 2*a*d

a = acceleration = (V^2 - U^2)/(2*d)

d = stopping distance = 1.7 cm = 0.017 m

V = 0 m/s, U = 31 km/hr = 31*5/18 = 8.61 m/s

So,

a = (0^2 - 8.61^2)/(2*0.017)

a = -2180.36 m/s^2

In two significant figures:

a = -2.2*10^3 m/s^2

Part B.

Since g = 9.81 m/s^2, So

a = -2180.36 m/s^2*(1 g/9.81 m/s^2) = (-2180.36/9.81)*g

a = -222.26*g

In two significant figures

a = -2.2*10^2g

Part C.

Using Newton's 2nd equation:

F_net = m*a

F_net = 67*(-2180.36) = -146084.12 N

In two significant figures:

F_net = -1.5*10^5 N

Don't forget to use negative sign before each answer.