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 $x = a + bt ^{2},$ where $a =8.5 m , b =2.5 ms ^{-2}$ and $t$ is measured in seconds.
The position of an object moving along X-axis is given by $x=a-b t^{2},$ where $a=8.5 m, b=2.5 ms ^{-2}$ 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 = 2.0 s $	1	$- 15 ms^{-1} $
B	Velocity of object at $t = 0s$	2	$- 10 ms^{-1} $
C	Instantaneous speed of object at$ t = 2.0 s $	3	$0 ms^{-1} $
D	Average velocity between $t = 2.0 s$ and $t = 4.0 s $	4	$ 10 ms^{-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, $x(t)=a-b t^{2}, a=8.5 m$ and $b=2.5 m / s ^{2}$
$=8.5-2.5 t^{2} $
Velocity of object $=\frac{ dx }{ dt }=-2 bt$
(A) velocity at $t=2.0 s =\left.\frac{ dx }{ dt }\right|_{t=2}=-4 b$
$=-4 \times 2.5=-10 ms ^{-1}$
(B) velocity at $\quad t =0=\left.\frac{ dx }{ dt }\right|_{t=0}=0 ms ^{-1}$
(C) Instantaneous speed = Magnitude of velocity
$=\left|-10 ms ^{-1}\right|=10 ms ^{-1}$
(D) Average velocity $=\frac{x\left(t_{2}\right)-x\left(t_{1}\right)}{t_{2}-t_{1}}$
$=\frac{x(4)-x(2)}{4-2}$
$=\frac{\left[a-b(4)^{2}\right]-\left[a-b(2)^{2}\right]}{2}$
$=\frac{4 b-16 b}{2}=-\frac{12 b}{2}=-6 b$
$=-6 \times 2.5\, ms ^{-1}=-15\, ms ^{-1}$