(i)     If $s = 100$ when $t = 0$ , find an expression for s in terms of a and t.

(ii)    If $s = 0$ when $t = 0$ , write down an expression for s in terms of a and t.

A train M slows down so that it comes to a stop at the station.

(i)     Find the time it takes train M to come to a stop, giving your answer in terms of a.

(ii)    Hence show that $a = \frac{8}{5}$ .

For a different train N, the value of a is 4.

Show that this train will stop before it reaches the station.

Note: In this question, do not penalize absence of units.

(i) $s = \int {(40 - at){\rm{d}}t}$     (M1)

$s = 40t - \frac{1}{2}a{t^2} + c$     (A1)(A1)

substituting $s = 100$ when $t = 0$ ($c = 100$ )     (M1)

$s = 40t - \frac{1}{2}a{t^2} + 100$     A1     N5

(ii) $s = 40t - \frac{1}{2}a{t^2}$     A1     N1

[6 marks]

(i) stops at station, so $v = 0$     (M1)

$t = \frac{{40}}{a}$ (seconds)     A1     N2

(ii) evidence of choosing formula for s from (a) (ii)     (M1)

substituting $t = \frac{{40}}{a}$     (M1)

e.g. $40 \times \frac{{40}}{a} - \frac{1}{2}a \times \frac{{{{40}^2}}}{{{a^2}}}$

setting up equation     M1

e.g. $500 = s$ , $500 = 40 \times \frac{{40}}{a} - \frac{1}{2}a \times \frac{{{{40}^2}}}{{{a^2}}}$ , $500 = \frac{{1600}}{a} - \frac{{800}}{a}$

evidence of simplification to an expression which obviously leads to $a = \frac{8}{5}$     A1

e.g. $500a = 800$ , $5 = \frac{8}{a}$ , $1000a = 3200 - 1600$

$a = \frac{8}{5}$     AG     N0

[6 marks]

METHOD 1

$v = 40 - 4t$ , stops when $v = 0$

$40 - 4t = 0$     (A1)

$t = 10$     A1

substituting into expression for s     M1

$s = 40 \times 10 - \frac{1}{2} \times 4 \times {10^2}$

$s = 200$     A1

since $200 < 500$ (allow FT on their s, if $s < 500$ )     R1

train stops before the station     AG     N0

METHOD 2

from (b) $t = \frac{{40}}{4} = 10$     A2

substituting into expression for s

e.g. $s = 40 \times 10 - \frac{1}{2} \times 4 \times {10^2}$     M1

$s = 200$     A1

since $200 < 500$      R1

train stops before the station     AG     N0

METHOD 3

a is deceleration     A2

$4 > \frac{8}{5}$    A1

so stops in shorter time     (A1)

so less distance travelled     R1

so stops before station     AG     N0

[5 marks]

Part (a) proved accessible for most.

Part (b), simple as it is, proved elusive as many candidates did not make the connection that $v = 0$ when the train stops. Instead, many attempted to find the value of t using $a = \frac{8}{5}$ .

Few were successful in part (c).