(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).