A car starts from rest and moves with uniform | vedclass

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(C) The motion can be represented by a velocity-time graph which is a triangle with base $2T$ and height $v_m$.For the first part ($t=0$ to $t=T$),the

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$\frac{aT}{2}$

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(C) The motion can be represented by a velocity-time graph which is a triangle with base $2T$ and height $v_m$.For the first part ($t=0$ to $t=T$),the car starts from rest with acceleration $a$. The maximum velocity reached is $v_m = aT$.The total distance $s$ traveled is equal to the area under the velocity-time graph.$s = 	ext{Area of triangle} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2T) 	imes v_m = T 	imes v_m$.Substituting $v_m = aT$,we get $s = T(aT) = aT^2$.The total time taken is $t_{total} = 2T$.The average speed is given by $v_{avg} = \frac{	ext{total distance}}{	ext{total time}} = \frac{aT^2}{2T} = \frac{aT}{2}$.

$A$ car starts from rest and moves with uniform acceleration $a$ on a straight road from time $t = 0$ to $t = T$. After that,a constant deceleration $a$ brings it to rest. In this process,the average speed of the car is

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