Constant acceleration with a velocity–time graph

Kinematics

A-level Maths (9MA0)

ORIGINAL

Sketch the velocity–time graph first: v = u + at for the speeds, the area under the graph for the distances.

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Question

A car moves along a straight horizontal road from

A

B

. It passes

A

with speed

4\text{ m s}^{-1}

and accelerates uniformly at

1.5\text{ m s}^{-2}

for 8 s. It then moves at constant speed for

T

s, and then decelerates uniformly, coming to rest at

B

10 s later. The distance

AB

is 400 m. (a) Find the speed of the car after the first 8 s. (b) Sketch a velocity–time graph for the motion. (c) Find

T

. (d) Find the deceleration.

Every step worked, with the reasoning.

1
$v = u + at = 4 + 1.5 \times 8 = 16\text{ m s}^{-1}$
Constant acceleration: use $v = u + at$ with $u = 4$ , $a = 1.5$ and $t = 8$ .
2
Graph of $v$ against $t$ : a straight line from $(0, 4)$ to $(8, 16)$ , horizontal to $(8 + T, 16)$ , then a straight line down to $(18 + T, 0)$ .
Constant acceleration gives straight-line sections; the velocity axis starts at 4, not 0.
3
Distance in the first 8 s $= \dfrac{4 + 16}{2} \times 8 = 80\text{ m}$
The area of the trapezium under the first section — the same as $s = \left(\frac{u + v}{2}\right)t$ .
4
Distance in the last 10 s $= \dfrac{1}{2} \times 10 \times 16 = 80\text{ m}$
The area of the triangle under the final section.
5
$80 + 16T + 80 = 400$ , so $16T = 240$ and $T = 15$
The total area under the graph is the distance $AB$ ; the middle rectangle has area $16T$ .
6
Deceleration $= \dfrac{16 - 0}{10} = 1.6\text{ m s}^{-2}$
The size of the gradient of the final section.

Answer: (a) $16\text{ m s}^{-1}$ (b) see the graph in step 2 (c) $T = 15$ (d) $1.6\text{ m s}^{-2}$