Figure gives the speed-time graph of the motion of a car. What is the ratio of the distance travelled by the car during the last two seconds to the total distance travelled in seven seconds?

Question

Figure gives the speed-time graph of the motion of a car. What is the ratio of the distance travelled by the car during the last two seconds to the total distance travelled in seven seconds? (A) (1/9) (B) (2/9) (C) (1/3) (D) (4/9)

Tardigrade Admin · Accepted Answer

Distance travelled in the last two seconds is

S =

area of triangle

B D C = \frac{1}{2} \times 20 \times 2 = 20 m

Total distance travelled in seven seconds is

S^{'} =

area of triangle

O A E +

area of rectangle

A B D E

+ area of triangle

B D C

= \frac{1}{2} \times 20 \times 3 + 20 \times 2 + \frac{1}{2} \times 20 \times 2 = 90 m

∴ \frac{S}{S ^{'}} = \frac{20 m}{90 m} = \frac{2}{9}

Q. Figure gives the speed-time graph of the motion of a car. What is the ratio of the distance travelled by the car during the last two seconds to the total distance travelled in seven seconds?

$\frac{1}{9}$

$\frac{2}{9}$

$\frac{1}{3}$

$\frac{4}{9}$

Q. Figure gives the speed-time graph of the motion of a car. What is the ratio of the distance travelled by the car during the last two seconds to the total distance travelled in seven seconds?

91​

92​

31​

94​

Distance travelled in the last two seconds is S= area of triangle BDC=21​×20×2=20m Total distance travelled in seven seconds is S′= area of triangle OAE+ area of rectangle ABDE + area of triangle BDC =21​×20×3+20×2+21​×20×2=90m ∴S′S​=90m20m​=92​

$\frac{1}{9}$

$\frac{2}{9}$

$\frac{1}{3}$

$\frac{4}{9}$