Your brother claims his car will accelerate from rest to a speed of 45 m/s in 8.36 seconds. assume the acceleration is constant.(a) what is the car’s acceleration? (b) how far does the car travel during that time period? (c) what is the speed of the car after 10.5 seconds (assuming it maintains the same acceleration)?

(a)  5.38 m/s²

(b) 188.004 m

(c) 56.49 m/s.

Explanation:

(a) Acceleration: This is the rate of change of velocity. The S.I unit of acceleration is m/s².

Acceleration can be represented as,

a = v-u/t Equation 1

Where a = acceleration, v = velocity, t = time

Given: v = 45 m/s, u = 0 m/s ( from rest)  t = 8.36 s.

Substituting into equation 1,

a = 45/8.36

a = 5.38 m/s²

Thus the acceleration of the car = 5.38 m/s²

(b) S = ut + 1/2at² Equation 2

Where S = distance, u = initial velocity, a = acceleration, t = time.

Given: u = 0 m/s (at rest), t = 8.36 s, a = 5.38 m/s²

Substituting into equation 2

S = 0(8.36) + 1/2(5.38)(8.36)²

S = 2.69(69.89)

S = 188.004 m

(c) Speed (v)

v = u + at Equation 3

Given: u = 0 m/s (at rest), a = 5.38 m/s², t = 10.5 s.

Substituting into equation 3

v = 0 + 5.38(10.5)

v = 56.49 m/s.

we need answer choices