M1 June 2013 Q4

EdexcelOld spec9 marksSUVAT

4. A lorry is moving along a straight horizontal road with constant acceleration. The lorry passes a point \(A\) with speed \(u\) m s\(^{-1}\), \((u \lt 34)\), and 10 seconds later passes a point \(B\) with speed 34 m s\(^{-1}\). Given that \(AB = 240\) m, find

(a) the value of \(u\), (3)

(b) the time taken for the lorry to move from \(A\) to the mid-point of \(AB\). (6)

Mark scheme (a)

Scheme	Marks
\(240 = \tfrac{1}{2}(u + 34)10\)	M1 A1
\(u = 14\)	A1
	(3)

Notes

First M1 for a complete method to produce an equation in \(u\) only.

First A1 for a correct equation. (\(u^2 - 48u + 476 = 0\) oe is possible).

Second A1 for \(u = 14\).

Mark scheme (b)

Scheme	Marks
\(34 = 14 + 10a\ \Rightarrow\ a = 2\)	M1 A1
\(120 = 14t + \tfrac{1}{2} \times 2 \times t^2\)	M1 A1
\(t^2 + 14t - 120 = 0\)
Solving, \(t = -20\) or 6	DM1
\(t = 6\)	A1
	(6)
	(9 marks)

OR

\(34 = 14 + 10a\ \Rightarrow\ a = 2\)	M1 A1
\(v^2 = 14^2 + 2 \times 2 \times 120\ \Rightarrow v = 26\)
AND \(26 = 14 + 2t\)	M1 A1
\(t = 6\)	DM1 A1
	(6)

Notes

EITHER

First M1 for an equation in \(a\) only. (M0 if \(v = 34\) when \(s = 120\) is used)

First A1 for \(a = 2\). (This may have been found in part (a))

Second M1 for a 3-term quadratic equation in \(t\) only, allow sign errors (must have found a value of \(a\). (M0 if \(v = 34\) when \(s = 120\) is used)

Second A1 for a correct equation.

Third M1 dependent on previous M1 for solving for \(t\).

Third A1 for \(t = 6\)

First M1 for an equation in \(a\) only.

First A1 for \(a = 2\). (This may have been found in part (a))

Second M1 for a complete method to obtain an equation in \(t\) only, allow sign errors. (must have found a value of \(a\))

Second A1 for a correct equation.

Third M1 dependent on previous M1 for solving for \(t\).

Third A1 for \(t = 6\)