The velocity-time graph of a car is given below. The car weighs $1000 \ kg$.
$(i)$ What is the distance travelled by the car in the first $2 \ s$?
$(ii)$ What is the braking force at the end of $5 \ s$ to bring the car to a stop within one second?

(N/A) The distance travelled by the car in the first $2 \ s$ is equal to the area under the velocity-time graph from $t = 0$ to $t = 2 \ s$,which is the area of $\Delta ABE$.
Distance $= \text{Area of } \Delta ABE = \frac{1}{2} \times \text{base} \times \text{height}$
$= \frac{1}{2} \times 2 \ s \times 15 \ m/s = 15 \ m$.
$(ii)$ To find the braking force,we first calculate the acceleration (deceleration) from $t = 5 \ s$ to $t = 6 \ s$ using the slope of the line $CD$.
Initial velocity at $t = 5 \ s$ is $u = 15 \ m/s$.
Final velocity at $t = 6 \ s$ is $v = 0 \ m/s$.
Time interval $\Delta t = 6 \ s - 5 \ s = 1 \ s$.
Acceleration $a = \frac{v - u}{\Delta t} = \frac{0 - 15 \ m/s}{1 \ s} = -15 \ m/s^2$.
Using Newton's second law of motion,$F = m \times a$.
$F = 1000 \ kg \times (-15 \ m/s^2) = -15000 \ N$.
The negative sign indicates that the force is a braking force acting opposite to the direction of motion. Thus,the magnitude of the braking force is $15000 \ N$.