Question

Two bodies P and Q have to move equal distances starting from rest. P is accelerated with 2a for first half distance then its acceleration becomes a for last half, where as Q has acceleration a for firsthalf and acceleration 2a for last half, then for whole journey.
(1) Average speed of P is more than that of Q.
(2) Average speed of both will be same. (3) Maximum speed during the journey is more for P.
(4) Maximum speed during the journey is more for Q.
Please explain answer​

Answer 1

Answer:

Average speed for P will be more during the journey.

Explanation:

Distance travelled by both P and Q is same.

Let us say the distance travelled by the is = x meter

for first half:

For P

distance travelled is x/2.

acceleration for first half is = 2a

then the time taken for first half distance from kinematic equation ($x=ut+\frac{1}{2}at^2$  )

u=0 for first half

$\frac{x}{2}= \frac{2a t_{1P}^2}{2}\\\implies t_{1P}^2= \frac{x}{2a}$

Similarly for Q in first half

$t_{1Q}=\frac{x}{a}$

$t_{1P}<t_{1Q}\\\implies {v_{1P}}>v_{1Q}$

for second half:

for P:

initial velocity will be v =u+at , u= 0 and acceleration was 2a hence v= 2at  ,

acceleration now is a , hence using the equation of kinematic

$\frac{x}{2}= 2at_{2P} ^2+ \frac{a}{2}t_{2P}^2\\\\\implies t_{2P}^2=\frac{x}{5a}$

Similarly for Q, acceleration for next half is 2a, and velocity is at

$\frac{x}{2}= at_{2Q}^2 + \frac{2a}{2}t_{2Q}^2\\\\\implies t_{2Q}^2= \frac{x}{4a}$

$\implies t_{2P}<t_{2Q}$

$\implies v_{2P}>v_{2Q}$

from the above calculation it can be seen that for first half of distance and second half of distance  velocity of P is greater than velocity of Q. Hence Average speed for P will be more during the journey