Two particles $P$ and $Q$ simultaneously start moving from point $A$ with velocities $15\,m/s$ and $20\,m/s$ respectively. The two particles move with accelerations equal in magnitude but opposite in direction. When $P$ overtakes $Q$ at $B$,then its velocity is $30\,m/s$. The velocity of $Q$ at point $B$ will be $..........\,m/s$.

The displacement $x$ of a particle depends on time $t$ as $x = \alpha t^2 - \beta t^3$.

At time $t=0$, a particle leaves the origin and moves in the positive direction of the $X$-axis. If the velocity of the particle varies as $v(t)=v_0(1-t/t_0)$, where $|v_0|=10 \ m/s$ and $t_0=10 \ s$, then the distance covered by the particle during the first $20 \ s$ is: (in $m$)

$A$ passenger arriving in a new town wishes to go from the station to a hotel located $10 \;km$ away on a straight road from the station. $A$ dishonest cabman takes him along a circuitous path $23 \;km$ long and reaches the hotel in $28 \;min$. What is
$(a)$ the average speed of the taxi,
$(b)$ the magnitude of average velocity? Are the two equal?

$A$ monkey climbs up a slippery pole for $3 \ s$ and subsequently slips for $3 \ s$. Its velocity at time $t$ is given by $v(t) = 2t(3 - t)$ for $0 < t < 3$ and $v(t) = -(t - 3)(6 - t)$ for $3 < t < 6$ in $m/s$. It repeats this cycle until it reaches the height of $20 \ m$.
$(a)$ At what time is its velocity maximum?
$(b)$ At what time is its average velocity maximum?
$(c)$ At what time is its acceleration maximum in magnitude?
$(d)$ How many cycles (counting fractions) are required to reach the top?

The equation of motion of a body is $\frac{dv}{dt} = -4v + 8$,where $v$ is the velocity in $m/s$ and $t$ is the time in seconds. The initial velocity of the particle was zero. Then,