The velocity-time graphs of a car and a scoot | vedclass

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Question

$\begin{array}{l}
{m{Using}}\,equation,\,a = \frac{{v - u}}{t}\,and\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

Vedclass · Accepted Answer

$112.5\,m$ અને $22.5\,s$

Vedclass · Answer

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

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

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