Question

The given plot shows the variation of acceleration ($m{s^{ - 2}}$) with time ($s$) for an object that started from rest at time $t{\text{ }} = {\text{ }}0{\text{ }}s$. The velocity at time $t{\text{ }} = {\text{ }}15{\text{ }}s$ (${v_{15}}$) and $25{\text{ }}s$ (${v_{25}}$) are

A. ${v_{15}}{\text{ }} = {\text{ }}50{\text{ }}m{s^{ - 1}}$ and ${v_{25}}{\text{ }} = {\text{ }}0{\text{ }}m{s^{ - 1}}\:$
B. ${v_{15}}{\text{ }} = {\text{ }}100{\text{ }}m{s^{ - 1}}$ and ${v_{25}}\; = {\text{ }}150{\text{ }}m{s^{ - 1}}\:$
C. ${v_{15}}{\text{ }} = {\text{ }}50{\text{ }}m{s^{ - 1}}$ and ${v_{25}}{\text{ }} = {\text{ }}25{\text{ }}m{s^{ - 1}}\:$
D. ${v_{15}}{\text{ }} = {\text{ }}100{\text{ }}m{s^{ - 1}}$ and ${v_{25}}{\text{ }} = {\text{ }}25{\text{ }}m{s^{ - 1}}\:$
E. None of these

Answer 1

Hint:We will use our basic concepts of motion graphs and then use the concepts for solving the given question. We can use the concept of distance or displacement being the area under the curve of a velocity-time graph. Finally, we will put the values in suitable formulae and then select the correct answer.

Formulae Used:
$\text{Area of triangle}= {\text{ }}\dfrac{1}{2}{\text{ }} \times {\text{ }}base{\text{ }} \times {\text{ }}{\left( {height} \right)^2}$
And
\[\text{Area of rectangle}= \text{length} \times \text{breadth}\]

Complete step by step answer:
We are given here with the acceleration time graph and we know that the area under the acceleration time graph gives us the velocity. Now, for the first time stamp of $t{\text{ }} = {\text{ }}15{\text{ }}s$, we will evaluate the velocity ${v_{15}}$. Now, area up to the timestamp of $15{\text{ }}s$ can be broken down in two parts area of the triangle from timestamp of $0{\text{ }}s$ to $10{\text{ }}s$ where base is \[10{\text{ }} - {\text{ }}0{\text{ }} = {\text{ }}10{\text{ }}units\], height is $10{\text{ }} - {\text{ }}0{\text{ }} = {\text{ }}10{\text{ }}units$.

Thus, the area is $ar{\text{ }}(10{\text{ }}s){\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }} \times {\text{ }}10{\text{ }} \times {\text{ }}10{\text{ }} = {\text{ }}50$.
And then the area of the rectangle from timestamp of $10{\text{ }}s$ to $15{\text{ }}s$ where the length is $10{\text{ }} - {\text{ }}0{\text{ }} = {\text{ }}10{\text{ }}units$, breadth is $15{\text{ }} - {\text{ }}10{\text{ }} = {\text{ }}5{\text{ }}units$.
Thus, area is $ar{\text{ }}(10{\text{ }}to{\text{ }}15){\text{ }} = {\text{ }}10{\text{ }} \times {\text{ }}5{\text{ }} = 50$.
Thus, the velocity of timestamp of $t{\text{ }} = {\text{ }}15{\text{ }}s$ is ${v_{15}}{\text{ }} = {\text{ }}ar{\text{ }}(10{\text{ }}s){\text{ }} + {\text{ }}ar{\text{ }}(10{\text{ }}to{\text{ }}15){\text{ }} = {\text{ }}50{\text{ }} + {\text{ }}50{\text{ }} = {\text{ }}100{\text{ }}m{s^{ - 1}}$.

Now, for the timestamp of $t{\text{ }} = {\text{ 2}}5{\text{ }}s$, we will evaluate the velocity ${v_{25}}$.Now, area up to the timestamp of $25{\text{ }}s$ can be broken down in three parts area of the triangle from timestamp of $0{\text{ }}s$ to $10{\text{ }}s$ where base is \[10{\text{ }} - {\text{ }}0{\text{ }} = {\text{ }}10{\text{ }}units\], height is $10{\text{ }} - {\text{ }}0{\text{ }} = {\text{ }}10{\text{ }}units$.
Thus, the area is $ar{\text{ }}(10{\text{ }}s){\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }} \times {\text{ }}10{\text{ }} \times {\text{ }}10{\text{ }} = {\text{ }}50$.
And then the area of the rectangle from timestamp of $10{\text{ }}s$ to $15{\text{ }}s$ where the length is $10{\text{ }} - {\text{ }}0{\text{ }} = {\text{ }}10{\text{ }}units$, breadth is $15{\text{ }} - {\text{ }}10{\text{ }} = {\text{ }}5{\text{ }}units$.

Thus, area is $ar{\text{ }}(10{\text{ }}to{\text{ }}15){\text{ }} = {\text{ }}10{\text{ }} \times {\text{ }}5{\text{ }} = 50$.
And, finally the area of the triangle from the timestamp of $15{\text{ }}s$ to $25{\text{ }}s$ with base $25{\text{ }} - {\text{ }}15{\text{ }} = {\text{ }}10{\text{ }}units$ and height of $10{\text{ }} - {\text{ }}0{\text{ }} = {\text{ }}10{\text{ }}units$.
Thus, the area of the triangle is
$ar{\text{ }}(15{\text{ }}to{\text{ }}25){\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }} \times {\text{ }}10{\text{ }} \times {\text{ }}10{\text{ }} = {\text{ }}50$
Thus, the velocity of timestamp of $t{\text{ }} = {\text{ 2}}5{\text{ }}s$ is ${v_{25}}{\text{ }} = {\text{ }}ar{\text{ }}(10{\text{ }}s){\text{ }} + {\text{ }}ar{\text{ }}(10{\text{ }}to{\text{ }}15){\text{ }} + {\text{ }}ar{\text{ }}(15{\text{ }}to{\text{ }}25){\text{ }} = {\text{ }}50{\text{ }} + {\text{ }}50{\text{ }} + {\text{ }}50{\text{ }} = {\text{ }}150{\text{ }}m{s^{ - 1}}$

Hence, the correct answer is B.

Note:Students often make mistakes while breaking the area into proper parts.Students should very carefully understand what is asked to find and then proceed to save time in channelizing in an incorrect direction. Students break down into parts where they fall into a situation of clumsy calculations.