Question

THE SPEED OF THE SCOOTER MOVING WITH UNIFORM ACCELERATION OF
4 M/S^2 BECOMES 20 M/S AT A CERTAIN TIME. WHAT WILL BE SPEED OF THE SCOOTER, WHEN IT HAS COVERED A DISTANCE OF 112M AFTER THAT TIME?
HINT V^2-U^2=2AX

Answer 1

Answer: It will be 36 m/s

Explanation:

Given

a=4m/s^2

u=20m/s

S=112m

Substituting the values in equation v2-u2=2as,

We get V=36m/s

Thank you

Answer 2

Answer:

$v=36 m/s$

Explanation:

We Have,

                $a=4m/s^2,\\u=20m/s,\\s=112m.$

Usiing the the third equation of motion,

$=>v^2-u^2=2as\\=>v^2-(20m/s)^2=2*4m/s^2*112m\\=>v^2-400m^2/s^2=896m^2/s^2\\=>v^2=(896+400)m^2/s^2\\=>v^2=1296m^2/s^2\\=>v=\sqrt{1296m^2/s^2}\\ =>v=36m/s$

So, Velocity after 112m = 36m/s.

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