Question

The speed-time graph of a car is given alongside. The car weighs 100 kg.



(i) What is the distance travelled by car in the first two seconds?
​(ii) What is the braking force applied at the end of 5 seconds to bring the car to a stop within one second?

Answer 1

(i) Area of the speed-time graph gives the total distance travelled.

In the first two seconds, the graph covers an area of a right-angled triangle with a height of 15 units and a base of 2 units.


So, the distance travelled by car in the first two seconds = Area of the triangle formed = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}=\frac{1}{2}\times 2\times 15=15\mathrm{m}$

Hence, the car travels 15 m in the first two seconds.

(ii) The slope of the speed-time graph at any moment of time gives the value of the acceleration.

The slope of the graph at the end of the 5 seconds = $\frac{15-0}{6-5}=15$

So, acceleration at the end of the 5 seconds = 15 m/s²

The braking force applied = Mass of the car $\times$ Acceleration of the car = $100\mathrm{kg}\times 15\mathrm{m}/{\mathrm{s}}^2=1500\mathrm{N}$

Hence, the braking force applied at the end of the 5 seconds is equal to 1500 N.