A-LevelMathematicsMechanics9709/42 · Oct/Nov 2023 · Q712 Marks

A particle X travels in a straight line. The velocity of X at time ts after leaving a fixed point O is denoted by vms⁻¹, where v = −0.1t³ + 1.8t² – 6t + 5.6. The acceleration of X is zero at t = p and t = q, where p < q.

(a)Find the value of p and the value of q.[4]
(b)It is given that the velocity of X is zero at t = 14. Find the velocities of X at t = p and at t = q, and hence sketch the velocity-time graph for the motion of X for 0 ≤ t ≤ 15.[3]
(c)Find the total distance travelled by X between t = 0 and t = 15.[5]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

[M1] Attempt to differentiate v a=(dv/dt) = 3x-0.1t^3-1 +2×1.8t^2–1 – 6t^1–1 = −0.3t² +3.6t-6 Decrease power by 1 and a change in coefficient in at least one term (which must be the same term); allow unsimplified; allow p or q for t. v/t is M0.
[A1] May be unsimplified. Allow p or q for t.
[DM1] Setting a = dv/dt = 0 and attempt to solve a 3 term quadratic for t. [a=dv/dt-=0⇒3t² –36t +60=0⇒t² –12t + 20 = 0] Allow p or q for t. Must get 2 values or numerical expressions for t from their three term quadratic. If using quadratic formula, must be the correct formula. If factorising, when brackets expanded, 2 terms correct.
[A1] p = 2, q=10
[B1] Velocities are 0ms⁻¹ and 25.6ms⁻¹ SOI
[B1] Curve with single minimum turning point followed by single maximum turning point Ignore placement of graph on axes Not a cusp for the minimum point or maximum point.
[DB1] All correct in 1st and 4th quadrant. Must go from convex to concave. Need to label 2 and 14 on the t-axis where the curve meets the t-axis. Do not need to show exact velocities at t = 0 or t = 10 or t = 15. Ignore graph outside 0≤t≤15.
[M1] Attempt to integrate v (s=) -0.1t^4/4 + 1.8t^3/3 – 6t^2/2 + 5.6t(+c) = -0.025t^4 +0.6t^3 -3t^2 +5.6t [+c] Increase power by 1 and a change in coefficient in at least one term (which must be the same term); s = vt is M0.
[A1] May be unsimplified.
[DM1] Attempt distance from t=0 to t=14 [= 176.4] Correct use of limits 0 and 14 for their s, i.e. F(14) - F(0) May see limits 0 to 2 and 2 to 14 used but must be (F(14)-F(2))+(F(2)-F(0)).
[DM1] Attempt distance from t=14 to t =15 [= (-) 8.025] Correct use of limits 14 and 15 for their s, i.e. ±(F(15)-F(14)). For reference F(2) = 18/5 =3.6, F(14) = 882/5 =176.4 and F(15) = 1347/8 =168.375.
[A1] Total distance =176.4+8.025=184.425 m = 7377/40 www Condone 184 or better.