The speed- time graph of a car is given in (figure) The car weights `1000kg`
(a) What is the distance travelled by the car in first two seconds?
(b) What is the braking force applied at the end of `5 seconds` to bring the car to a stop within one second?

Explanation

(i) We will first calculate the distance travelled in the first two seconds. From the given graph we find that :
Distance travelled in first two seconds = Area under line AB and the time axis
=Area of triangle AB2
`=1/2xx`base`xx`Height
`=1/2xx2xx15`
=15 m
Thus, the distance travelled by the car in the first two seconds is 15 metres.
(ii) The braking force is to be calculated by using the formula
Force=mass`xx`acceleration
or F=m`xx`a
Here, the mass of the car is given as 1000 kg but the acceleration in the last one second (from point C to point D) is to be obtained from the graph shown above. Now, we know that the acceleration is given by the slope of speed-time graph. So,
Acceleration in last one second = Slope of line CD
=`("Perpendicular")/("Base")`
`=15/1`
`=15m//s^(2)`
If we look at the above given graph, we will find that the speed of the car is decreasing from point C to point D. That is, the acceleration from C to D is negative and hence it should be written with a minus sign. Thus,
Acceleration, a =-15 `m//s^(2)`
Now,putting m= 1000 kg and a =15 m/`s^(2)` in formula (1), we get:
Force,F=1000 `xx` (-15)
=-15000N
Thus, the force applied by the brakes to stop the car is 15000 newtons. The negative sign of force here shows that the force is being applied in a direction opposite to the motion of the car. That is, it is a retarding force.