Question

Given questions are based on following passage. Choose the correct option from those given below.
The position of an object moving along $X$ -axis is given by $x=a+b{t}^2,$ where $a=8.5m,b=2.5m{s}^{-2}$ and $t$ is measured in seconds.
The position of an object moving along X-axis is given by $x=a-b{t}^2,$ where $a=8.5m,b=2.5m{s}^{-2}$ and $t$ is measured in seconds.
For the above situation, match the terms in Column I with the values of Column II and choose the correct option from the codes given below:

Column I		Column II
A	Velocity of object at $t=2.0s$	1	$-15m{s}^{-1}$
B	Velocity of object at $t=0s$	2	$-10m{s}^{-1}$
C	Instantaneous speed of object at$t=2.0s$	3	$0m{s}^{-1}$
D	Average velocity between $t=2.0s$ and $t=4.0s$	4	$10m{s}^{-1}$

• A. $A-1,B-2,C-3,D-4$
• B. $A-2,B-3,C-4,D-1$
• C. $A-4,B-3,C-2,D-1$
• D. $A-3,B-2,C-1,D-4$

Answer 1

Answer: (B)

Given, $x(t)=a-b{t}^2,a=8.5m$ and $b=2.5m/s^2$
$=8.5-2.5t^2$
Velocity of object $=\frac{dx}{dt}=-2bt$
(A) velocity at $t=2.0s={\frac{dx}{dt}|}_{t=2}=-4b$
$=-4\times 2.5=-10m{s}^{-1}$
(B) velocity at $t=0={\frac{dx}{dt}|}_{t=0}=0m{s}^{-1}$
(C) Instantaneous speed = Magnitude of velocity
$=\left|-10m{s}^{-1}\right|=10m{s}^{-1}$
(D) Average velocity $=\frac{x\left(t_2\right)-x\left(t_1\right)}{t_2-t_1}$
$=\frac{x(4)-x(2)}{4-2}$
$=\frac{\left[a-b(4{)}^2\right]-\left[a-b(2{)}^2\right]}{2}$
$=\frac{4b-16b}{2}=-\frac{12b}{2}=-6b$
$=-6\times 2.5m{s}^{-1}=-15m{s}^{-1}$