On entering a state highway, a car continues to move with a constant velocity of 6 ms⁻¹ for 2 minutes and then accelerates with a constant acceleration 1 ms⁻² for 6 seconds. Find the displacement of the car on the state highway in the 2 min 6 s time interval by drawing a velocity-time graph for its motion.

The car moves with a constant velocity of 6 ms$^{-1}$ for 2 minutes and then accelerates for 6 seconds.

The required velocity-time graph:

We know that

$$\begin{aligned} v &= u + at \\ v &= 6 + (1 \times 6) \\ v &= 12 \text{ ms}^{-1} \end{aligned}$$

The displacement is equal to the total area under the velocity-time graph.

Displacement during first 120 s:

$$\begin{aligned} s_1 &= 6 \times 120 \\ s_1 &= 720 \text{ m} \end{aligned}$$

Displacement during next 6 s:

$$\begin{aligned} s_2 &= \frac{(6+12)}{2} \times 6 \\ s_2 &= 9 \times 6 \\ s_2 &= 54 \text{ m} \end{aligned}$$

Therefore,

Total displacement = 720 + 54

$$= 774 \text{ m}$$

The displacement of the car in 2 min 6 s is 774 m.