Question

A vehicle initially at rest covers a distance of 128m in 8 sec from the point of start . calculate the acceleration and final velocity

Answer 1

Body is at rest

This means that :-

Initial velocity = 0m/sec

$\rule{200}{2}$

Distance covered = 128m

Time taken= 8 sec

From the second Equation of Motion we have :-

$s=ut+\dfrac{1}{2}at^2\\ \\ \\128 = 0*8 +\dfrac{1}{2}a*64\\ \\ \\128=\dfrac{64}{2}a\\ \\ \\a=\dfrac{128*2}{64}\\ \\ \\a=2*2\\ \\ \\a=4 m/sec^2$

$\rule{200}{2}$

From the First equation of Motion we have :-

$v=u+at\\ \\ \\v=0+4(8)\\ \\ \\v=32m/sec$

$\rule{200}{2}$

ACCELERATION = 4m/sec^2

FINAL VELOCITY = 32 m/sec

Answer 2

Answer:

MOTION :

Vehicle was initially at rest,

That means initial velocity will be 0.

Initial velocity, u = 0 m/s

Distance covered, s = 128 m

Time taken, t = 8 s

Using 2nd Equation of motion,

$\boxed{\mathsf{s\:=\:ut\:+\:{\dfrac{1}{2}* a{t} ^{2}}}}$

$\mathsf{128 \:=\: 0*t\: +\: {\dfrac{1}{2}*a{8} ^{2}}}$

$\mathsf{128 \:=\: 32a}$

$\mathsf{a\:=\:{\dfrac{128}{32}}}$

⛛ $\mathsf{a\:=\:4 \: m{s} ^{-2}}$

Now, Using 1st Equation of Motion,

$\boxed{\mathsf{v \:=\:u\:+\:at}}$

$\mathsf{v\:=\:0\:+\:4*8}$

⛛ $\mathsf{v\:=\:32\:m{s}^{-1}}$