12 exam-style questions on constant acceleration, SUVAT, velocity–time graphs and variable acceleration. Tap Show answer for full working.
A particle moves in a straight line with constant acceleration \(2\,\mathrm{m\,s}^{-2}\). It starts from rest. Find its velocity and the distance travelled after \(5\,\mathrm{s}\).
Using \(v = u + at\): \(v = 0 + 2\cdot5 = 10\,\mathrm{m\,s}^{-1}\).
Using \(s = ut + \dfrac{1}{2}at^2\): \(s = 0 + \dfrac{1}{2}\cdot2\cdot25 = 25\,\mathrm{m}\).
A car accelerates from \(10\,\mathrm{m\,s}^{-1}\) to \(30\,\mathrm{m\,s}^{-1}\) over a distance of \(200\,\mathrm{m}\) with constant acceleration. Find the acceleration and the time taken.
\(v^2 = u^2 + 2as \Rightarrow 30^2 = 10^2 + 2a\cdot200 \Rightarrow 900 = 100 + 400a \Rightarrow a = 2\,\mathrm{m\,s}^{-2}\).
\(v = u + at \Rightarrow 30 = 10 + 2t \Rightarrow t = 10\,\mathrm{s}\).
The velocity–time graph of a particle consists of a straight line from \((0,0)\) to \((4,12)\), then a straight line to \((10,0)\). Find the total distance travelled and the average speed over the \(10\,\mathrm{s}\).
Area under graph: triangle \(0\)–\(4\): \(\dfrac{1}{2}\cdot4\cdot12 = 24\). Triangle \(4\)–\(10\): \(\dfrac{1}{2}\cdot6\cdot12 = 36\). Total distance \(= 60\,\mathrm{m}\).
Average speed \(= \dfrac{60}{10} = 6\,\mathrm{m\,s}^{-1}\).
A ball is thrown vertically upwards with speed \(14\,\mathrm{m\,s}^{-1}\). Taking \(g = 9.8\,\mathrm{m\,s}^{-2}\) and neglecting air resistance, find the maximum height reached and the time until it returns to the point of projection.
At max height \(v=0\): \(0 = 14^2 + 2(-9.8)s \Rightarrow 0 = 196 - 19.6s \Rightarrow s = 10\,\mathrm{m}\).
Time to return: \(s=0\): \(0 = 14t - 4.9t^2 \Rightarrow t(14-4.9t)=0 \Rightarrow t=0\) or \(t=\dfrac{14}{4.9}\approx 2.86\,\mathrm{s}\).
A particle travels in a straight line. Its velocity \(v\,\mathrm{m\,s}^{-1}\) at time \(t\,\mathrm{s}\) is given by \(v = 3t^2 - 12t + 9\) for \(0\le t\le 5\).
(a) Find the times when the particle is instantaneously at rest.
(b) Find the total distance travelled in the first \(5\) seconds.
(a) \(v=0\): \(3t^2-12t+9=0 \Rightarrow t^2-4t+3=0 \Rightarrow (t-1)(t-3)=0\). Rest at \(t=1\) and \(t=3\).
(b) Displacement \(s = \int v\,\mathrm{d}t = t^3 - 6t^2 + 9t + C\). At \(t=0\), \(s=0\), so \(C=0\).
From \(0\) to \(1\): \(s(1)=1-6+9=4\). From \(1\) to \(3\): \(s(3)=27-54+27=0\), so displacement \(-4\). From \(3\) to \(5\): \(s(5)=125-150+45=20\), displacement \(20\).
Total distance \(= 4 + 4 + 20 = 28\,\mathrm{m}\).
A train accelerates uniformly from rest to \(25\,\mathrm{m\,s}^{-1}\) in \(50\,\mathrm{s}\), travels at constant speed for \(2\) minutes, then decelerates uniformly to rest in \(40\,\mathrm{s}\). Find the total distance travelled and the average speed for the whole journey.
Acceleration phase: \(s_1 = \dfrac{1}{2}\cdot25\cdot50 = 625\,\mathrm{m}\).
Constant speed: \(s_2 = 25\cdot120 = 3000\,\mathrm{m}\).
Deceleration: \(s_3 = \dfrac{1}{2}\cdot25\cdot40 = 500\,\mathrm{m}\).
Total distance \(= 625+3000+500 = 4125\,\mathrm{m}\).
Total time \(= 50+120+40 = 210\,\mathrm{s}\). Average speed \(= \dfrac{4125}{210} = 19.64\,\mathrm{m\,s}^{-1}\) (3 s.f.).
A particle moves along a straight line. Its acceleration at time \(t\) seconds is \(a = 6t - 4\,\mathrm{m\,s}^{-2}\). When \(t=0\), the particle is at the origin with velocity \(2\,\mathrm{m\,s}^{-1}\).
(a) Find expressions for the velocity and displacement in terms of \(t\).
(b) Find the maximum displacement from the origin in the interval \(0\le t\le 3\).
(a) \(v = \int(6t-4)\,\mathrm{d}t = 3t^2 - 4t + C\). \(t=0\), \(v=2 \Rightarrow C=2\). So \(v = 3t^2 - 4t + 2\).
\(s = \int v\,\mathrm{d}t = t^3 - 2t^2 + 2t + D\). \(t=0\), \(s=0 \Rightarrow D=0\). So \(s = t^3 - 2t^2 + 2t\).
(b) Stationary points: \(v=0\): \(3t^2-4t+2=0\). Discriminant \(16-24=-8<0\), so \(v>0\) always. Max displacement in interval is at \(t=3\): \(s(3)=27-18+6=15\,\mathrm{m}\).
Two particles \(A\) and \(B\) start from the same point at the same time and move along the same straight line. \(A\) moves with constant velocity \(8\,\mathrm{m\,s}^{-1}\). \(B\) starts from rest and accelerates at \(2\,\mathrm{m\,s}^{-2}\). Find the time when \(B\) overtakes \(A\) and the distance from the start at that instant.
Displacement of \(A\): \(s_A = 8t\). Displacement of \(B\): \(s_B = \dfrac{1}{2}\cdot2\cdot t^2 = t^2\).
Overtake when \(s_B = s_A\): \(t^2 = 8t \Rightarrow t(t-8)=0\). \(t=8\,\mathrm{s}\) (ignoring \(t=0\)).
Distance \(= 8\cdot8 = 64\,\mathrm{m}\).
A particle moves in a straight line with velocity \(v = t^3 - 6t^2 + 9t\) for \(t\ge 0\).
(a) Find the times at which the acceleration is zero and determine the nature of each stationary point of the velocity.
(b) Sketch the velocity–time graph for \(0\le t\le 5\), marking the intercepts and stationary points.
(c) Find the total distance travelled between \(t=0\) and \(t=5\).
(a) \(a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = 3t^2 - 12t + 9 = 3(t^2-4t+3)=3(t-1)(t-3)\). Zero at \(t=1,3\).
Second derivative \(6t-12\): at \(t=1\), \(-6<0\) (local max of \(v\)); at \(t=3\), \(6>0\) (local min of \(v\)).
(b) \(v=t(t-3)^2\). Zeros at \(t=0,3\). Local max at \(t=1\): \(v=1\cdot4=4\). Local min at \(t=3\): \(v=0\). At \(t=5\), \(v=5\cdot4=20\).
(c) \(s=\int_0^5 v\,\mathrm{d}t\). \(v=t(t^2-6t+9)=t^3-6t^2+9t\). Antiderivative \(\dfrac{1}{4}t^4-2t^3+\dfrac{9}{2}t^2\).
From \(0\) to \(3\): \(\dfrac{1}{4}\cdot81 - 2\cdot27 + \dfrac{9}{2}\cdot9 = 20.25 - 54 + 40.5 = 6.75\). From \(3\) to \(5\): value at \(5\) minus value at \(3\): \((\dfrac{1}{4}\cdot625 - 2\cdot125 + \dfrac{9}{2}\cdot25) - 6.75 = (156.25-250+112.5)-6.75=18.75-6.75=12\). Total distance \(=6.75+12=18.75\,\mathrm{m}\).
A car is travelling at \(20\,\mathrm{m\,s}^{-1}\) when the driver sees a hazard \(60\,\mathrm{m}\) ahead. The reaction time of the driver is \(0.6\,\mathrm{s}\), after which the brakes produce a constant deceleration of \(5\,\mathrm{m\,s}^{-2}\). Does the car stop before hitting the hazard? Find the distance by which it stops short or the speed at impact.
During reaction: distance \(=20\cdot0.6=12\,\mathrm{m}\). Remaining distance to hazard \(=48\,\mathrm{m}\).
Braking: \(v^2=u^2+2as \Rightarrow 0=400+2(-5)s \Rightarrow 10s=400 \Rightarrow s=40\,\mathrm{m}\).
Stops after \(40\,\mathrm{m}\) of braking, total distance from seeing hazard \(=12+40=52\,\mathrm{m}<60\,\mathrm{m}\). Stops \(8\,\mathrm{m}\) short.
A particle starts from rest at the origin and moves along the \(x\)-axis with acceleration \(a = 2 + 3t\,\mathrm{m\,s}^{-2}\) for the first \(4\) seconds. For \(t>4\) the acceleration is constant and equal to the value it has at \(t=4\).
(a) Find the velocity and position at \(t=4\).
(b) Find the position when \(t=10\).
(a) For \(0\le t\le4\): \(v=\int(2+3t)\,\mathrm{d}t=2t+\dfrac{3}{2}t^2\) (since \(v(0)=0\)). At \(t=4\): \(v=8+24=32\,\mathrm{m\,s}^{-1}\).
\(s=\int v\,\mathrm{d}t=t^2+\dfrac{1}{2}t^3\) ( \(s(0)=0\)). At \(t=4\): \(s=16+32=48\,\mathrm{m}\).
(b) For \(t>4\), \(a=2+3\cdot4=14\,\mathrm{m\,s}^{-2}\) constant. From \(t=4\) to \(t=10\) (\(\Delta t=6\)): \(\Delta v=14\cdot6=84\), so \(v(10)=32+84=116\).
\(\Delta s = 32\cdot6 + \dfrac{1}{2}\cdot14\cdot36 = 192 + 252 = 444\). Position \(=48+444=492\,\mathrm{m}\).
The velocity of a particle at time \(t\) seconds is \(v = 4\cos 2t - 3\sin 2t\,\mathrm{m\,s}^{-1}\), \(t\ge0\). The particle starts at the origin.
(a) Show that the acceleration is \(a = -8\sin 2t - 6\cos 2t\).
(b) Find the maximum speed of the particle.
(c) Find the displacement from the origin when the particle first comes to instantaneous rest.
(a) \(a=\dfrac{\mathrm{d}v}{\mathrm{d}t}=-8\sin 2t - 6\cos 2t\).
(b) \(v^2 = (4\cos 2t-3\sin 2t)^2 = 16\cos^2 + 9\sin^2 -24\sin\cos\). Max of \(R\cos(2t+\alpha)\) form: write \(v=R\cos(2t+\alpha)\) where \(R\cos\alpha=4\), \(R\sin\alpha=3\), so \(R=5\). Max speed \(=5\,\mathrm{m\,s}^{-1}\).
(c) Rest when \(v=0\): \(4\cos 2t = 3\sin 2t \Rightarrow \tan 2t = \dfrac{4}{3}\). First positive: \(2t=\tan^{-1}(4/3)\approx0.927\), \(t\approx0.464\,\mathrm{s}\).
\(s=\int v\,\mathrm{d}t = 2\sin 2t + \dfrac{3}{2}\cos 2t + C\). \(t=0\), \(s=0\): \(0=0+\dfrac{3}{2}+C\Rightarrow C=-\dfrac{3}{2}\). So \(s=2\sin 2t + \dfrac{3}{2}\cos 2t - \dfrac{3}{2}\).
At rest: \(\sin 2t=\dfrac{4}{5}\), \(\cos 2t=\dfrac{3}{5}\) (first quadrant). \(s=2\cdot\dfrac{4}{5} + \dfrac{3}{2}\cdot\dfrac{3}{5} - \dfrac{3}{2} = \dfrac{8}{5} + \dfrac{9}{10} - \dfrac{3}{2} = 1.6+0.9-1.5=1\,\mathrm{m}\).