Question

For the velocity-time graph shown in figure below, the distance covered by the body in the last two seconds of its motion is what fraction of the total distance covered in the seven seconds?

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B

$\frac{1}{4}$

Step 1: Given and to find

Velocity-time graph is given and needs to find the distance covered by the body in the last two seconds of its motion is what fraction of the total distance covered in the seven seconds.

Step 2: Finding the total distance covered in the seven seconds:

1. We know that Area under the velocity-time graph gives the distance covered by the body.
2. The area of the tringle= the distance travelled in the last two seconds.
3. Area of tringle =$\frac{1}{2}\times b\times h$ where b is base and h is height.

Then distance travelled in the last two seconds = $\frac{1}{2}\times (7-5)\times 10=\frac{1}{2}\times 2\times 10=10m^2$

Then the total distance covered in all seven seconds = the trapezium's area.

Distance covered in all seven seconds = $\frac{1}{2}\times \left[\right(7-1)+2]\times \left(10\right)=40m^2$

Now, divide the total distance travelled by the fraction of distance covered in the last two seconds. = $\frac{10m^2}{40m^2}=\frac{1}{4}$

Hence, the correct option is (B) $\frac{1}{4}$