Question

A sports car accelerates uniformly from rest to a speed of 87 mi/hr in 8s. Determine: a.The acceleration of the car

Answer 1

Answer:

Part a)

$a = 4.86 m/s^2$

Part b)

$d = 155.52 m$

Part c)

$v_f = 48.6 m/s$

Explanation:

As we know that car start from rest and reach to final speed of 87 mph

so we have

$v_f = 87 mph = 38.88 m/s$

now we have

Part a)

acceleration is rate of change in velocity

$a = \frac{v_f - v_i}{t}$

$a = \frac{38.88 - 0}{8}$

$a = 4.86 m/s^2$

Part b)

distance moved by car with uniform acceleration is given as

$d = \frac{v_f + v_i}{2} t$

$d = \frac{38.88 + 0}{2} 8$

$d = 155.52 m$

Part c)

As we know that the car start from rest

so final speed after t = 10 s

$v_f = v_i + at$

$v_f = 0 + (4.86)10$

$v_f = 48.6 m/s$