Name the physical quantity which corresponds to the rate of change of momentum.

State whether the following statement is true or false:
$A$ quantity which can be represented completely by magnitude alone is called a vector quantity.

The following table shows the position of three persons between $8.00\, am$ to $8.20\, am$.

Time	Position of Person $A$ $(km)$	Position of Person $B$ $(km)$	Position of Person $C$ $(km)$
$8.00\, am$	$0$	$0$	$0$
$8.05\, am$	$4$	$5$	$10$
$8.10\, am$	$13$	$10$	$19$
$8.15\, am$	$20$	$15$	$24$
$8.20\, am$	$25$	$20$	$27$

$(i)$ Who is moving with constant speed?
$(ii)$ Who has travelled the maximum distance between $8.00\, am$ to $8.05\, am$?
$(iii)$ Calculate the average speed of person $A$ in $km\, h^{-1}$.

$(a)$ Draw velocity-time graphs for the following cases:
$(i)$ When the object is at rest.
$(ii)$ When the object is thrown vertically upwards.
$(b)$ $A$ motorcyclist riding motorcycle $A$ who is travelling at $36 \, km \, h^{-1}$ applies the brakes and stops the motorcycle in $10 \, s$. Another motorcyclist of motorcycle $B$ who is travelling at $18 \, km \, h^{-1}$ applies the brakes and stops the motorcycle in $20 \, s$. Plot the speed-time graph for the two motorcycles. Which of the two motorcycles travelled farther before it came to a stop?

$A$ car is moving on a straight road with uniform acceleration. The following table gives the speed of the car at various instants of time.

Time $(s)$	$0$	$10$	$20$	$30$	$40$	$50$
Speed $(m s^{-1})$	$5$	$10$	$15$	$20$	$25$	$30$

$(i)$ Draw the speed-time graph representing the above set of observations.
$(ii)$ Find the acceleration of the car.

$A$ train starting from rest picks up a speed of $10\, m s^{-1}$ in $100\, s$. It continues to move at the same speed for the next $250\, s$. It is then brought to rest in the next $50\, s$. Plot a speed-time graph for the entire motion of the train and calculate:
$(i)$ acceleration of the train while accelerating,
$(ii)$ retardation of the train while retarding,
$(iii)$ the total distance covered by the train.