Question

A scooter starts from rest, moves in a straight line with a constant acceleration and covers a distance of 64 m in 4s.
(a) calculate its acceleration &final velocity​

Answer 1

Given:-

Initial velocity(u)=0

Distance covered(s)=64m

Time(t)=4s

To find:-

i.acceleration(a)

ii.final velocity(v)

Solution:-

By,using the second equation of motion,we get------

=>s=ut+1/2at²

=>64=0×4+1/2×a×(4)²

=>64=8a

=>a=64/8

=>a=8m/s²

Now,by using the first equation of motion,we get------

=>v=u+at

=>v=0+8×4

=>v=32m/s

Hence,acceleration is 8m/s² and final velocity is 32m/s.

Answer 2

Given:

Initial velocity = 0 m/s

Acceleration = a m/ s2

Time = t =4 s

Distance = s=64m

Find V=? And a=?

From Second equation of motion, we have:

S= ut + 1/2 at^2

64=0×t + (1/2) a(4×4)

64=16a/2

64=8a

a=64/8

a=8 m/s2

From FIRST equation of motion:

V=U+at

=0+8×4

V=32m/s

Therefore , Final velocity = 32m/s and acceleration is 8m/s2

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