Question

A particle is thrown in vertically in upward direction and passes three equally speed windows of equal highest. Then

• A. The average speed of the particle while passing the windows satisfy the relation $v_{av1}>v_{av2}>v_{av3}$
• B. The time taken by the particle to cross the windows satisfies the relation $t_1<t_2<t_3$
• C. The magnitude of the acceleration of the particle while crossing the windows, satisfies the relation ${\alpha }_1={\alpha }_2\ne {\alpha }_3$
• D. The change in the speed of the particle, while crossing the windows, would satisfy the relation $△v_1<△v_2<△v_3$

Answer 1

Answer: (A), (B), (D)

The correct options are
A The average speed of the particle while passing the windows satisfy the relation $v_{av1}>v_{av2}>v_{av3}$
B The time taken by the particle to cross the windows satisfies the relation $t_1<t_2<t_3$
D The change in the speed of the particle, while crossing the windows, would satisfy the relation $△v_1<△v_2<△v_3$

As the particle is going up, it is slowing down, i.e., speed is decreasing and hence we can say that time taken by the particle to cover equal distances is increasing as the particle is going up.
Hence, $t_1<t_2<t_3$.
As $v_{av}=\frac{Distance}{Time}$ , we have $v_{av}\propto \frac{1}{Time}$
So, $v_{av1}>v_{av2}>v_{av3}$
Acceleration throughout the motion remains same from equation,
$\overrightarrow{v}=\overrightarrow{u}+\overline{at},\Delta v\propto t$ So $\Delta v_1<\Delta v_2<\Delta v_3$