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

Question

For the velocity-time graph shown in figure below the distance covered by the body in last two seconds of its motion is what fraction of the total distance covered by it in all the six seconds? (A) (1/2) (B) (1/4) (C) (1/3) (D) (2/3)

Tardigrade Admin · Accepted Answer

Distance covered = Area enclosed by velocity-time graph.
Total distance covered in 6 seconds,

S = \frac{1}{2} \times 2 \times 10 + 2 \times 10 + \frac{1}{2} \times 2 \times 10 = 40 m

Distance covered in last two seconds,

S^{'} = \frac{1}{2} \times 2 \times 10 = 10 m

∴

Fraction of distance

= \frac{S ^{'}}{S} = \frac{10}{40} = \frac{1}{4}

Q. For the velocity-time graph shown in figure below the distance covered by the body in last two seconds of its motion is what fraction of the total distance covered by it in all the six seconds?

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{3}$

$\frac{2}{3}$

Q. For the velocity-time graph shown in figure below the distance covered by the body in last two seconds of its motion is what fraction of the total distance covered by it in all the six seconds?

21​

41​

31​

32​

Distance covered = Area enclosed by velocity-time graph. Total distance covered in 6 seconds, S=21​×2×10+2×10+21​×2×10=40m Distance covered in last two seconds, S′=21​×2×10=10m ∴ Fraction of distance =SS′​=4010​=41​

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{3}$

$\frac{2}{3}$