Question

A particle moves on a straight line and its a - t graph is shown.
List - I gives some quantites and List - II gives some possible magnitudes of quantites then correct match is

$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \text{I) Average velocity of 4 sec}& \text{P) 0}\\ \text{II) Average acceleration of 4 sec}& \text{Q) 5}\\ \text{III) Displacement in 4 sec}& \text{R) 10}\\ \text{IV) Velocity at t = 4 sec}& \text{S) 20}\\ & \text{T) 40}\\ & \text{U) 50}\end{array}$

• A. $I\to R,II\to Q,III\to T,IV\to S$
• B. $I\to S,II\to R,III\to T,IV\to U$
• C. $I\to R,II\to Q,III\to U,IV\to T$
• D. $I\to R,II\to P,III\to T,IV\to S$

Answer 1

The correct option is A $I\to R,II\to Q,III\to T,IV\to S$

Average acceleration, $<a>=\frac{\Delta v}{\Delta t}=\frac{\text{Area under a - t graph}}{\Delta }=\frac{\frac{1}{2}\times 4\times 10}{4}=5m/s^2$
From 0 to 2 secs,
$a=5t$
${\int }_0^{v}dv={\int }_0^{t}5tdtv=\frac{5}{2}{t}^{2}v\left(2\right)=\frac{5}{2}\times 2^2=10m/sand{\int }_0^{x_1}dx={\int }_0^2\frac{5}{2}{t}^{2}dt{x}_1=\left[\frac{5}{2}\right(\frac{t^3}{3}){]}_0^2=\frac{20}{3}$
From 2 to 4 sec,
$a=20-5t$
${\int }_0^{v}dv={\int }_2^t(20-5t)dtv=20t-\frac{5t^2}{2}+catt=2s,v=10\Rightarrow 10=40-\frac{5}{2}\times 2^2+c\Rightarrow c=-20v=20t-\frac{5t^2}{2}-20{\int }_0^{x_2}dx={\int }_2^4(20t-\frac{5t^2}{2}-20)dt{x}_2=\frac{100}{3}$

$\therefore$ Displacement $=x_1+x_2=\frac{20}{3}+\frac{100}{3}=\frac{120}{3}=40m<v>=\frac{\text{Displacement}}{time}=\frac{40}{4}=10m/s$
Say velocity at t = 4 is $v_4$.
Then $v_4-0$ = Area of a - t graph $v=\frac{1}{2}\times 4\times 10=20m/s$