If a car covers $2/5^{th}$ of the total distance with speed $v_1$ and $3/5^{th}$ of the total distance with speed $v_2$,then the average speed is:

The engine of a motorcycle can produce a maximum acceleration $5 \, m/s^2$. Its brakes can produce a maximum retardation $10 \, m/s^2$. What is the minimum time in which it can cover a distance of $1.5 \, km$?

$A$ particle starts moving from time $t=0$ and its coordinate is given as $x(t)=4t^{3}-3t.$ Which of the following statements are correct?
$A$. The particle returns to its original position (origin) at $t = \frac{\sqrt{3}}{2} \approx 0.866$ units later.
$B$. The particle is $1$ unit away from the origin at its turning point.
$C$. Acceleration of the particle is non-negative for $t \ge 0$.
$D$. The particle is $0.5$ units away from the origin at its turning point.
$E$. The particle never turns back as acceleration is non-negative.

$A$ horse rider covers half the distance with $5\,m/s$ speed. The remaining part of the distance was travelled with speed $10\,m/s$ for half the time and with speed $15\,m/s$ for the other half of the time. The mean speed of the rider averaged over the whole time of motion is $\frac{x}{7}\,m/s$. The value of $x$ is

$A$ body is projected vertically up with a velocity $v$ and after some time it returns to the point from which it was projected. The average velocity and average speed of the body for the total time of flight are

$A$ particle moves along a straight line such that its displacement $x$ varies with time $t$ as $x = \alpha t^3 + \beta t^2 + \gamma$,where $\alpha, \beta, \gamma$ are constants. $V_1$ is the average velocity of the particle during its journey between $t = 1 \ s$ and $t = 3 \ s$. $V_2$ is the instantaneous velocity of the particle at $t = 3 \ s$. The ratio $\frac{V_1}{V_2}$ is