Question

A body starts from rest and moves with an acceleration of 10 m/sq. s and attains a velocity of 50 m/s. Calculate the distance travelled by the body during this time.​

Answer 1

Provided that:

• Initial velocity = 0 mps
• Acceleration = 10 mps sq.
• Final velocity = 50 mps

To calculate:

• Distance travelled

Solution:

• Distance travelled = 125 m

Using concept:

• Third equation of motion

Using formula:

• ${\small{\underline{\boxed{\pmb{\sf{v^2 \: - u^2 \: = 2as}}}}}}$

Where, a denotes acceleration, u denotes initial velocity, v denotes final velocity and s denotes displacement or distance or height.

Required solution:

${\sf{:\implies v^2 \: - u^2 \: = 2as}}$

${\sf{:\implies (50)^{2} - (0)^{2} = 2(10)(s)}}$

${\sf{:\implies 2500 - 0 = 2(10)(s)}}$

${\sf{:\implies 2500 = 2(10)(s)}}$

${\sf{:\implies 2500 = 20s}}$

${\sf{:\implies \dfrac{2500}{20} = s}}$

${\sf{:\implies \dfrac{250}{2} = s}}$

${\sf{:\implies 125 = s}}$

${\sf{:\implies s \: = 125 \: m}}$

${\sf{:\implies Distance \: = 125 \: m}}$

Know more:

$\begin{gathered}\boxed{\begin{array}{c}\\ {\pmb{\sf{Three \: equation \: of \: motion}}} \\ \\ \sf \star \: v \: = u \: + at \\ \\ \sf \star \: s \: = ut + \: \dfrac{1}{2} \: at^2 \\ \\ \sf \star \: v^2 - u^2 \: = 2as\end{array}}\end{gathered}$

Answer 2

• initial velocity=u=0m/s
• Final velocity=v=50m/s
• Acceleration=10m/s^2
• Time=t

Using Acceleration formula

$\\ \sf\longmapsto a=\dfrac{v-u}{t}$

$\\ \sf\longmapsto t=\dfrac{v-u}{a}$

$\\ \sf\longmapsto t=\dfrac{50-0}{10}$

$\\ \sf\longmapsto t=\dfrac{50}{10}$

$\\ \sf\longmapsto t=5s$

• Distance=s

Using second equation of kinematics

$\boxed{\sf s=ut+\dfrac{1}{2}at^2}$

$\\ \sf\longmapsto s=0(5)+\dfrac{1}{2}(10)(5)^2$

$\\ \sf\longmapsto s=5(5)^2$

$\\ \sf\longmapsto s=5(25)$

$\\ \sf\longmapsto s=125m$