Question

Calculate acceleration of an object moving from a stop to 21 m/s in 3.9 s

Answer 1

5.38
21 divided by 3.9 is 5.38 and the equation for acceleration is change in velocity divided by time

Answer 2

Heya!!

For calculate aceleration, lets applicate formula:

                                                   $\boxed{a=\frac{V-V_o}{t} }$

                                                   Δ   Being   Δ

                                          V = Final Velocity = 21 m/s

                                         Vo = Initial Velocity = 0 m/s

                                          a = Aceleration =  ? m/s²

                                                  t = Time = 3,9 s

⇒ Let's replace according the formula:

$\boxed{a=\frac{21\ m/s - 0 \ m/s}{3,9\ s} }$

⇒ Resolving

$\boxed{ a=5,38\ m/s^{2}}$

Result:

The aceleration of the object is 5,38 m/s²

Good Luck!!