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

12.50 m/s

Step-by-step 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

Answer 2

• From first equation of motion:

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

substitute:

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