Velocity-time graphs of a car and a scooter are shown in figure. (i) difference between distance tra

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2.Motion in Straight Line

hard

The velocity-time graphs of a car and a scooter are shown in the figure. $(i)$ the difference between the distance travelled by the car and the scooter in $15\, s$ and $(ii)$ the time at which the car will catch up with the scooter are, respectively

A$337.5\,m$ and $25\,s$

B$225.5\,m$ and $10\,s$

C$112.5\,m$ and $22.5\,s$

D$112.5\,m$ and $15\,s$

(JEE MAIN-2018)

Solution

$\begin{array}{l}
{\rm{Using}}\,equation,\,a = \frac{{v – u}}{t}\,and\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,s = ut + \frac{1}{2}a{t^2}\\
{\rm{Distance}}\,{\rm{travelled}}\,{\rm{by}}\,{\rm{car}}\,{\rm{in}}\,{\rm{15}}\,{\rm{sec}}\\
{\rm{ = }}\frac{1}{2}\frac{{\left( {45} \right)}}{{15}}{\left( {15} \right)^2}\\
= \frac{{675}}{2}\,m\\
{\rm{Distance}}\,{\rm{traveled}}\,{\rm{by}}\,{\rm{scooter}}\,{\rm{in}}\,{\rm{15}}\,{\rm{seconds}}
\end{array}$
$\begin{array}{l}
{\rm{ = 30}} \times 15 = 450\left( {{\rm{distance}} = speed \times time} \right)\\
Difference\,between\,{\rm{distance}}\,travelled\,by\,car\\
and\,scooter\,in\,15\,\sec ,\,450 – 337.5 = 112.5\,m\\
Let\,car\,catches\,scooter\,in\,time\,t;\\
\frac{{675}}{2} + 45\left( {t – 15} \right) = 30t\\
337.5 + 45t – 675 = 30t\\
\Rightarrow \,15t = 337.5\\
\Rightarrow \,t = 22.5\,\sec
\end{array}$

Standard 11

Physics

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hard

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medium

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hard

(JEE MAIN-2021)

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medium

A body starts with an initial velocity of $10\,ms ^{-1}$ and is moving along a straight line with constant acceleration. When the velocity of the particle is $50\,ms ^{-1}$, the acceleration is reversed in direction. The velocity of the particle when it again reaches the starting point is $…………\,m/s^{-1}$

hard

The velocity-time graphs of a car and a scooter are shown in the figure. $(i)$ the difference between the distance travelled by the car and the scooter in $15\, s$ and $(ii)$ the time at which the car will catch up with the scooter are, respectively

Solution

Similar Questions

A train accelerates from rest at a constant rate $\alpha$ for distance $x_1$ and time $t_1$. After that it retards to rest at constant rate $\beta$ for distance $x_2$ and time $t_2$. Which of the following relations is correct?

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