Question

Given questions are based on the given information. There are four options for following questions out of which only one is correct. The motion of a car is represented by the equation.
$x(t)=0.08t^3$
Where, $x(t)$ is position of the car at time instant 't' There is a table given below providing the value of $\Delta x/\Delta t$ calculated for $\Delta t$ equals $2.0s,1.0s,0.5s$
$0.1s$ and $0.01s$ centred at $t=4.0s$. The second and
the third columns give the value of $t_1=\left(t-\frac{\Delta t}{2}\right)$ and $t_2=\left(t+\frac{\Delta t}{2}\right)$ and the fourth and fifth columns give the
corresponding values of ' $x$ '.

$\Delta t(s)$	$t_1(s)$	$t_2(s)$	$x(t)(m)$	$\Delta x(m)$	$\Delta x/\Delta t(m{s}^{-1})$
2.0	3.0	5.0	2.16	7.84	3.92
1.0	3.5	4.5	3.43	B	3.86
0.5	3.75	A	4.21875	1.9225	3.84
0.1	3.95	4.05	4.93039	0.38402	3.8402
0.01	3.995	4.005	5.100824	0.0384	C

The value of $\frac{dx}{dt}$ at $t=4.0s$

• A. The instantaneous velocity at $t=4.0s$
• B. The average velocity for $\Delta t=0.01$ s
• C. Same at the value of $C$
• D. All of the above

Answer 1

Answer: (D)

For given expression, for motion is
$x(t)=0.08t^3\Rightarrow \frac{dx}{dt}=0.24t^2$
${\frac{dx}{dt}|}_{t=4}=0.24\times (4{)}^2=0.24\times 16$
$=3.8400m{s}^{-1}$
We know instantaneous velocity $=v=\frac{dx}{dt}$
Also, from the table we observe that for $\Delta t=0.01s$, the
value of $\frac{\Delta x}{\Delta t}=3.8400$ which is same as instantaneous
velocity at $t=4s$