Kinematics of a particle in a straight line

Constant acceleration equations and speed-time and distance-time graphs.

A train starts from rest at station $P$ . It accelerates uniformly for $30\,\text{s}$ until it reaches a speed of $V\,\text{m s}^{-1}$ . It then travels at this constant speed for $100\,\text{s}$ , and finally decelerates uniformly for $20\,\text{s}$ , coming to rest at station $Q$ . The distance from $P$ to $Q$ is $2.5\,\text{km}$ .

(a) Using a speed–time graph, or otherwise, find the value of $V$ .

(b) Find the acceleration of the train during the first $30\,\text{s}$ .

(a) M1 Distance = area under speed–time graph, a trapezium with parallel sides $150$ and $100$ (1)
M1 $\tfrac{1}{2}(150 + 100)V = 2500$ (1)

A small ball is projected vertically upwards with speed $19.6\,\text{m s}^{-1}$ from a point $A$ which is $24.5\,\text{m}$ above horizontal ground. The ball is modelled as a particle moving freely under gravity.

(a) Find the time taken for the ball to reach the ground.

(b) Find the speed of the ball as it hits the ground.

(a) M1 Taking up as positive: $-24.5 = 19.6t - 4.9t^2$ (1)
A1 $t^2 - 4t - 5 = 0$ , so $t = 5\,\text{s}$ (1)
(b) $v = 19.6 - 9.8 \times 5 = -29.4$ , so speed $29.4\,\text{m s}^{-1}$ (1)
(c) M1 $14.7 = 19.6t - 4.9t^2$ (1)
A1 $t^2 - 4t + 3 = 0$ , so $t = 1$ or $t = 3$ (1)
A1 Time at least $14.7\,\text{m}$ above $A$ is $3 - 1 = 2\,\text{s}$ (1)

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