Question

A physics student running 2.8 m/s increases her velocity to 4.7 m/s in 3.4 seconds. What is the magnitude of acceleration of the student?

Answer 1

Answer:

The magnitude of acceleration of the student is 0.6 m/s²

Explanation:

Given that,

Initial velocity = 2.8 m/s

Final velocity = 4.7 m/s

Time = 3.4 sec

We need to calculate the magnitude of acceleration of the student

Using formula of acceleration

$a=\dfrac{v_{f}-v_{i}}{t}$

Where, $v_{f}$ = final velocity

$v_{f}$ = final velocity

t = time

Put the value into the formula

$a=\dfrac{4.7-2.8}{3.4}$

$a=0.6\ m/s^2$

Hence, The magnitude of acceleration of the student is 0.6 m/s²