A car starts from rest and travels with uniform acceleration $\alpha$ for some time and then with uniform retardation $\beta$ and comes to rest. If the total travel time of the car is $‘t’$, the maximum velocity attained by it is given by :-
$\frac{\alpha \beta}{(\alpha + \beta)}.t$
$\frac{1}{2} \frac{\alpha \beta}{(\alpha + \beta)}.t^2$
$\frac{\alpha \beta}{(\alpha - \beta)}.t$
$\frac{1}{2} \frac{\alpha \beta}{(\alpha - \beta)}.t^2$
Acceleration versus velocity graph of a particle moving in a straight line starting from rest is as shown in figure. The corresponding velocity-time graph would be
A very large number of balls are thrown vertically upwards in quick succession in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is $5\, m$, the number of balls thrown per minute is (take $g = 10\, ms^{-2}$) :-
The displacement-time graph for two particles $A$ and $B$ are straight lines inclined at angles of $30^o$ and $60^o$ with the time axis. The ratio of velocities of $V_A : V_B$ is
The maximum possible acceleration of a train moving on a straight track is $10\ m/s^2$ and maximum possible retardation is $5\ m/s^2$ . If maximum achievable speed of train is $10\ m/s$ then minimum time in which train can complete a journey of $135\ m$ starting from rest and ending at rest, is .........$s$
The motion of a particle along a straight line is described by equation $x = 8 + 12t -t^3$ where $x$ is in metre and $t$ in second. The retardation of the particle when its velocity becomes zero is.......$ms^{-2}$