Question

if a car starting from rest is accelerated at the rate of 0.2m/s2 find its find velocity at the end of 1 minute. what will be the distance covered during this time?​

Answer 1

Explanation:

given

U=0

A=0.2m/s^2

V=?

T=1 minutes

S=?

find final velocity first

Apply first equation of motion

V=U+at

V=0+0.2×1

V=0.2

Now distance covered

S=ut +1/2+at^2

S=0×1+1/2+0.2×1×1

S=1/2+0.2×1

S=0.1m

Answer 2

$\underline{ \underline{ \large\green\maltese\bf \large \red{Given:-}}}$

$\bull$ Acceleration,a = $0.2 m/s^2$

$\bull$ Time Period,t = 1 minute = 60 s

$\bull$Initial Velocity,u =0

$\:\:\;\:\:\;\:\:\;\:\:\:\:\;\:\:\;(Body\;was\;at\;rest\;initially)$

$\underline{ \underline{ \large\purple\maltese\bf \large \orange{To \:Find:-}}}$

$\bull$Distance covered during given time period.

$\underline{ \underline{ \large\blue\maltese\bf \large \green{Solution:-}}}$

$\underline{ \underline{ \sf \large \purple{Formula \: Used - }}}$

$\boxed{s = ut + \frac{1}{2} a {t}^{2} }$

Where

• S= Distance
• u= Initial Velocity
• a = acceleration
• t = time period

Putting Values,

${S = ut + \frac{1}{2} a {t}^{2} }$

${S = 0×t+ \frac{1}{2} ×0.2× {60}^{2} }$

${S = \frac{1}{2} ×0.2× 3600 }$

${S = \frac{1}{\cancel2} ×\cancel{720} }$

$\boxed{S = 360 \;m}$

$\sf\red{\dag Distance \;covered \;during \:the\; time=360 m}$