15.4 The constant-acceleration formulae — Practise

Work through these on your own. Write down the three quantities you know and the one you want. Choose the formula that omits the quantity you were not given. Keep one positive direction for the whole question.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

A cyclist moves at $6$ m/s and accelerates at $3$ m/s $^2$ for $5$ s. Find the final velocity and the displacement.
A car starts from rest and accelerates at $4$ m/s $^2$ for $6$ s. Find its final velocity and the distance travelled.
A car travelling at $25$ m/s brakes uniformly and stops in $100$ m. Find the acceleration and the time taken.
A particle starts from rest and accelerates at $2$ m/s $^2$ over a distance of $36$ m. Find its final velocity and the time taken.
A ball is thrown vertically upwards at $14$ m/s. Find the greatest height it reaches. Take $g = 9.8$ m/s $^2$ .
A ball is thrown vertically upwards at $19.6$ m/s. Find the time it takes to reach its greatest height. Take $g = 9.8$ m/s $^2$ .
A ball is dropped from rest and falls for $3$ s. Taking downwards as positive, find its final velocity and the distance fallen. Take $g = 9.8$ m/s $^2$ .
A car travelling at $30$ m/s brakes at $6$ m/s $^2$ until it stops. Find the stopping time and the stopping distance.
A train accelerates uniformly from $12$ m/s to $24$ m/s over $8$ s. Find its acceleration and the distance travelled.
A stone is thrown vertically upwards at $9.8$ m/s and returns to the thrower's hand at the same height. Find the total time in the air and the greatest height. Take $g = 9.8$ m/s $^2$ .
A particle moving at $5$ m/s accelerates at $2$ m/s $^2$ over a distance of $44$ m. Find its final velocity.
A car starts from rest and accelerates at $2$ m/s $^2$ for $5$ s. It then travels at a constant velocity for $10$ s. Find the total distance travelled.

Answers and concise solutions

1. The final velocity and the displacement

Answer: Use $v = u + at$ : $v = 6 + 3 \times 5 = 21$ m/s. Then $s = ut + \frac{1}{2}at^2 = 30 + 37.5 = 67.5$ m.

2. The final velocity and the distance

Answer: From rest, $u = 0$ : $v = 0 + 4 \times 6 = 24$ m/s, and $s = \frac{1}{2} \times 4 \times 36 = 72$ m.

3. The acceleration and the time

Answer: Use $v^2 = u^2 + 2as$ with $v = 0$ : $0 = 625 + 200a$ , so $a = -\dfrac{625}{200} = -3.125$ m/s $^2$ . The average velocity is $12.5$ m/s, so $t = \dfrac{100}{12.5} = 8$ s.

4. The final velocity and the time

Answer: Use $v^2 = u^2 + 2as$ : $v^2 = 0 + 2 \times 2 \times 36 = 144$ , so $v = 12$ m/s. Then $v = u + at$ gives $12 = 2t$ , so $t = 6$ s.

5. The greatest height

Answer: At the greatest height $v = 0$ . Use $v^2 = u^2 + 2as$ with $a = -9.8$ : $0 = 196 - 19.6s$ , so $s = 10$ m.

6. The time to the greatest height

Answer: At the greatest height $v = 0$ . Use $v = u + at$ with $a = -9.8$ : $0 = 19.6 - 9.8t$ , so $t = 2$ s.

7. The final velocity and the distance

Answer: With downwards positive, $u = 0$ and $a = 9.8$ : $v = 9.8 \times 3 = 29.4$ m/s, and $s = \frac{1}{2} \times 9.8 \times 9 = 44.1$ m.

8. The stopping time and the distance

Answer: The car stops, so $v = 0$ . From $v = u + at$ : $0 = 30 - 6t$ , so $t = 5$ s. Then $s = ut + \frac{1}{2}at^2 = 150 - 75 = 75$ m.

9. The acceleration and the distance

Answer: Use $a = \dfrac{v - u}{t} = \dfrac{24 - 12}{8} = 1.5$ m/s $^2$ . Then $s = \dfrac{u + v}{2}t = \dfrac{12 + 24}{2} \times 8 = 144$ m.

10. The total time and the greatest height

Answer: The time to the greatest height is $t = \dfrac{u}{g} = 1$ s, so the total time is $2$ s. The greatest height is $\dfrac{u^2}{2g} = \dfrac{9.8^2}{19.6} = 4.9$ m.

11. The final velocity

Answer: Use $v^2 = u^2 + 2as$ : $v^2 = 25 + 176 = 201$ , so $v = \sqrt{201} \approx 14.2$ m/s. The value is not a whole number, so the surd form is the exact answer.

12. The total distance

Answer: In the first stage the car travels $\frac{1}{2} \times 2 \times 25 = 25$ m and reaches $10$ m/s. In the second stage it travels $10 \times 10 = 100$ m. The total is $125$ m.