Question

A 2700kg truck is moving at 9.2m/s. A 320N braking force is applied to the truck. How far will it travel by the time it comes to a stop?

Answer 1

Answer:

$\huge\boxed{\sf S = 357.075\ m}$

Explanation:

Given data:

Mass = m = 2700 kg

Force = F = -320 N (Negative because it is braking force)

Initial velocity = $V_i$ = 9.2 m/s

Final velocity = $V_f$ = 0 m/s

Required:

Distance = S = ?

Solution:

First,

Finding a (acceleration):

We know that,

F = ma

Put the given data.

-320 = 2700 (a)

Divide both sides by 2700

-320/2700 = a

a ≈ -0.1185 m/s²

Now,

Finding distance:

$2aS=v_f^2-v_i^2$ (Third equation of motion)

Put the given data

2(-0.1185)S = (0) - (9.2)²

-0.237 (S) = - 84.64

• Divide both sides by -0.237

S = -84.64 / -0.237

S = 357.075 m

$\rule[225]{225}{2}$