Question

A car traveling initially at 9.49 m/s accelerates at the rate of 0.988 m/s^2 for 3.05s. What is it’s velocity at the end of the acceleration?

Answer 1

Answer:

• From first equation of motion:

${ \boxed{ \bf{v = u + at}}}$

substitute:

$v = 9.49 + (0.988 \times 3.05) \\ v = 9.49 + 3.0134 \\ v = 12.50 \: \: m {s}^{ - 1}$

Answer 2

Answer:

12.50 m/s

Explanation:

Vi = 9.49 m/s

a = 0.988 m/s²

t = 3.05 s

Vf = ?

Vf = Vi + at

Vf = 9.49 + (0.988)(3.05)

Vf = 12.50 m/s