Question

A particle moves in
straight line with a constant acceleration.It changes its velocity from 10m/s to 20m/s while passing through
a distance of 135m in t seconds. The value of t is______​

Answer 1

Answer:

A particle moves in  straight line with a constant acceleration . It changes its velocity from 10m/s to 20m/s while passing through  a distance of 135m in t seconds

To Find :

Value of t

[ As the acceleration is constant throughout the motion in a straight line , we need to apply equations of kinematics  \orange{\bigstar}★ ]

Solution :

\begin{gathered}\bigstar\ \; \sf \purple{v^2-u^2=2as}\\\\\bigstar\ \; \sf \red{v=u+at}\end{gathered}★ v2−u2=2as★ v=u+at

where ,

v denotes final velocity

u denotes initial velocity

a denotes acceleration

s denotes displacement / distance

t denotes time

Given ,

Initial velocity , u = 10 m/s

Final velocity , v = 20 m/s

Distance , s = 135 m

Note : In a straight line , distance and displacement are both same

\begin{gathered}\bigstar\ \; \pink{\sf v^2-u^2=2as}\\\\\to \sf (20)^2-(10)^2=2a(135)\\\\\to \sf 400-100=270a\\\\\to \sf 300=270a\\\\\to \sf a=\dfrac{300}{270}\\\\\to \sf a=\dfrac{30}{27}\\\\\leadsto \sf a=\dfrac{10}{9}\ m/s^2\ \; \bigstar\end{gathered}★ v2−u2=2as→(20)2−(10)2=2a(135)→400−100=270a→300=270a→a=270300→a=2730⇝a=910 m/s2 ★

\begin{gathered}\bigstar\ \; \sf \blue{v=u+at}\\\\\to \sf 20=10+(\dfrac{10}{9})t\\\\\to \sf 20-10=\dfrac{10t}{9}\\\\\to \sf 10=\dfrac{10t}{9}\\\\\to \sf 90=10t\\\\\to \sf 10t=90\\\\\leadsto \sf \orange{t=9\ s}\ \; \bigstar\end{gathered}★ v=u+at→20=10+(910)t→20−10=910t→10=910t→90=10t→10t=90⇝t=9 s ★

Answer 2

Answer:

i hope it will help u alot