Question

a runner accelerates at a rate of 3.2 m/s^2 if the runner starts from 9.23m/s after 6.7s how fast are they going?

Answer 1

Given data:

* The acceleration of the runner is,

`a=3.2ms^(-2)`

* The initial velocity of the runner is,

`u=9.23ms^(-1)`

* The time given is,

`t=6.7s`

Solution:

By the kinematics equation, the final velocity of the runner is,

`v-u=at`

where v is the final velocity,

Substituting the known values,

`\begin{gathered} v-9.23=3.2*6.7 \\ v-9.23=21.44 \\ v=21.44+9.23 \\ v=30.67ms^(-1) \end{gathered}`

Thus, the final velocity of the runner is 30.67 m/s.