A 2290 kg car traveling to the west at 22.3 m/s slows down uniformly. How long would it take the car to come to a stop if the force on the car is 8930 N to the east? Let East be positive. Answer in units of s.

5.72 s

Explanation:

From Newton's law, F = ma

The East is +ve direction, Hence,

F = +8930 N

m = 2290 kg

a = ?

8930 = 2290 × a

a = 8930/2290 = 3.90 m/s²

So, we will find the time it takes the car to stop using the equations of motion

a = 3.90 m/s²

u = initial velocity of the car = - 22.3 m/s (the velocity is to the west)

v = final velocity of the car = 0 m/s (since the car comes to rest)

t = time taken for the car to come to rest = ?

v = u + at

0 = - 22.3 + (3.90)(t)

3.9t = 22.3

t = 5.72 s

5.72 s.

Explanation:

F = 8930 N

M = 2290 kg

u = 22.3 m/s

Note, F = M × a

a = 8930 ÷ 2290

= 3.9 m/s^2

Using equation of motion,

v = u + at

22.3 = 3.9t

t = 5.72 s.

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