A 975-kg pickup comes to rest from a speed of 87.5 km/h in a distance of 125 m. Suppose the pickup is initially traveling in the positive direction. (a) If the brakes are the only thing making the car come to a stop, calculate the force (in newtons, in a component along the direction of motion of the car) that the brakes apply on the car .
(b) Suppose instead of braking that the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force, in newtons, exerted on the car in this case 33%
(c) What is the ratio of the force on the car from the concrete to the braking force?

A) Force = 2303.925 N in the negative x-direction

B) F ≈ 143998.28 N

C) Ratio = 62.5

Explanation:

A) Since the brakes are the only thing making the van to come to a stop, then first of all, we will calculate the force (in a component along the direction of motion of the car) that the brakes will apply on the van.

Let's find the deceleration using Newton's law of motion formula;

v² = u² + 2as

where;

v = final velocity,

u = initial velocity,

s = displacement

a = acceleration

We are given;

u = 87.5 km/h = 24.3056 m/s

s = 125 m

v = 0 m/s

Thus;

0 = (24.3056)² + 2a(125)

- (24.3056)²= 250a

a = - 24.3056²/250

a = - 2.363 m/s²

Now, force = mass × acceleration

We are given mass = 975 kg

Thus;

Force = 975 x (-2.363)

Force = 2303.925 N in the negative x-direction

B) formula for kinetic energy is

KE = ½mv²

KE = ½(975)(24.3056)²

= 287996.568288 J

Work done on impact = F x 2

Thus;

2F = 287996.568288

F = 287996.568288/2

F ≈ 143998.28 N

C) Ratio = Force on car/braking force = 143998.284/2303.925 = 62.5

answer is c

explanation:

c. newton's third law of motion