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IB Mathematics: Applications and Interpretation HL · AHL 2: Further Derivatives · AHL 2.9
Kinematics: displacement, velocity and acceleration Problem set B One question at a time
Try each question on paper first. Open the working a line at a time, then say whether you had it.
Skills
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A bead slides along a straight wire. The displacement from a fixed point is \(s\) cm, where \(s=3 t^{3} - 10 t^{2} + 7 t + 2\) and \(t\ge0\) is in seconds.
(a) Find the velocity when \(t=2\).
- \(v=\frac{ds}{dt}=9 t^{2} - 20 t + 7\)
- \(v(2)=3\)
(b) Find the acceleration when \(t=2\).
- \(a=\frac{dv}{dt}=18 t - 20\)
- \(a(2)=16\)
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A raft drifts along a straight river. The displacement from a fixed point is \(s=20e^{-0.5t}+2t\) m, where \(t\ge0\) is in seconds. Give your answers to 3 significant figures.
(a) Find the velocity when \(t=1\).
- \(v=2-10e^{-0.5t}\)
- \(v(1)=−4.07\)
(b) Find the acceleration when \(t=1\).
- \(a=5e^{-0.5t}\)
- \(a(1)=3.03\)
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A sled slides along a straight slope. The displacement from a fixed point is \(s=2 t^{3} - 21 t^{2} + 60 t\) m, for \(t\ge0\) seconds.
(a) Find the times when the particle is at rest.
- \(v=6 t^{2} - 42 t + 60\)
- \(v=0\Rightarrow 6 \left(t - 5\right) \left(t - 2\right)=0\)
- \(t=2,\ t=5\)
(b) Find the displacement at each of those times.
- \(s(2)=52\)
- \(s(5)=25\)
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A weight bobs on an elastic string. The displacement from its rest position is \(x=5\cos(2t)\) cm, where \(t\) is in seconds. Use radians.
(a) Find the velocity when \(t=1\). Give your answer to 3 significant figures.
- \(v=\frac{dx}{dt}=- 10 \sin{\left(2 t \right)}\)
- \(v(1)=−9.09\)
(b) Find the first time after \(t=0\) when the object is at rest.
- \(v=0\Rightarrow \sin(2t)=0\Rightarrow 2t=\pi\)
- \(t=1.57\)
(c) Write down the greatest speed of the object.
- \(|v|\le 10\)
- The greatest speed is 10 cm s⁻¹.
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A cyclist rides along a straight road. Its displacement from O is \(s=t^{3} - 15 t^{2} + 63 t\) metres after \(t\) seconds, \(t\ge0\). For each time, say whether the particle is speeding up or slowing down.
(a) \(t=2\)
- \(v=3 t^{2} - 30 t + 63,\ a=6 t - 30\)
- \(v(2)=15,\ a(2)=−18\)
- v and a have opposite signs, so the particle is slowing down.
(b) \(t=4\)
- \(v=3 t^{2} - 30 t + 63,\ a=6 t - 30\)
- \(v(4)=−9,\ a(4)=−6\)
- v and a have the same sign, so the particle is speeding up.
(c) \(t=6\)
- \(v=3 t^{2} - 30 t + 63,\ a=6 t - 30\)
- \(v(6)=−9,\ a(6)=6\)
- v and a have opposite signs, so the particle is slowing down.
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A skater glides along a straight line of ice. Its acceleration is \(a=6 t - 10\) m s\(^{-2}\). When \(t=0\) the velocity is 7 m s\(^{-1}\) and the displacement is -1 m.
(a) Find an expression for the velocity.
- \(v=\int a\,dt=3 t^{2} - 10 t+c\)
- \(v(0)=7\Rightarrow c=7\)
(b) Find an expression for the displacement.
- \(s=\int v\,dt=t^{3} - 5 t^{2} + 7 t+k\)
- \(s(0)=-1\Rightarrow k=-1\)
(c) Find the displacement when \(t=3\).
- \(s(3)=2\)
In context
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A sledge slides along a flat snowfield after a push. Its distance from the start is \(s=45\left(1-e^{-0.15t}\right)\) metres after \(t\) seconds, \(t\ge0\).
(a) Find the velocity when \(t=4\). Give your answer to 3 significant figures.
- \(v=6.75e^{-0.15t}\)
- \(v(4)=3.70\)
(b) Find the time when the velocity is 0.5 m s\(^{-1}\). Give your answer to 3 significant figures.
- \(6.75e^{-0.15t}=0.5\)
- \(t=-\frac{1}{0.15}\ln\left(\frac{0.5}{6.75}\right)=17.4\)
(c) Find the distance that the sledge approaches and never passes.
- \(t\to\infty:\ e^{-0.15t}\to0,\ s\to45\)
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The top of an engine piston moves up and down. Its height above the ground is \(h=15-6\cos(1.2t)\) cm after \(t\) seconds, where \(t\ge0\). Use radians.
(a) Find the vertical velocity when \(t=2\). Give your answer to 3 significant figures.
- \(v=\frac{dh}{dt}=7.2\sin(1.2t)\)
- \(v(2)=4.86\)
(b) Find the greatest rate at which the piston rises or falls.
- The sine function has greatest value 1.
- The greatest rate is 7.2 cm s⁻¹.
(c) Find the first time that the piston is rising at 4 cm s\(^{-1}\). Give your answer to 3 significant figures.
- \(7.2\sin(1.2t)=4\)
- Solve the equation with the calculator (radian mode), or by taking the inverse sine.
- The first time is t = 0.491 s.
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A train moves along a straight section of line. Its velocity is \(v=t^{2} - 8 t + 12\) m s\(^{-1}\) for \(0\le t\le7\). Give your answers to 3 significant figures.
(a) Find the times when the velocity is zero.
- \(t^{2} - 8 t + 12=0\)
- \(t=2,\ t=6\)
(b) Find the displacement from the starting point after 7 seconds.
- \(\int_{0}^{7}v\,dt=2.33\)
- Forwards and backwards parts partly cancel.
(c) Find the total distance travelled in the 7 seconds.
- \(\int_{0}^{7}|v|\,dt=23.7\)
- The distance is 23.7 m.
Stretch
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A probe moves along a straight tube. Its displacement from O is \(s=8te^{-0.5t}\) metres after \(t\) seconds, \(t\ge0\). Give your answers to 3 significant figures where needed.
(a) Find expressions for the velocity and the acceleration.
- Product rule with \(u=8t\) and the exponential factor.
- \(v=8(1-0.5t)e^{-0.5t}\)
- \(a=4(0.5t-2)e^{-0.5t}\)
(b) Find the time when the particle is at rest, and its displacement then.
- \(v=0\Rightarrow 1-0.5t=0\Rightarrow t=2\)
- \(s(2)=5.89\)
(c) Find the time when the acceleration is zero, and the velocity then.
- \(a=0\Rightarrow 0.5t-2=0\Rightarrow t=4\)
- \(v(4)=−1.08\)
- The velocity is negative: the particle is moving back towards O.
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A drone moves along a straight line. Its velocity is \(v=6\sin(0.5t)-2.5\) m s\(^{-1}\) for \(0\le t\le9\). Use radians and give your answers to 3 significant figures.
(a) Find the times when the particle is at rest.
- \(6\sin(0.5t)=2.5\)
- \(t=0.860\text{ and }t=5.42\)
- Use the calculator to find both solutions in the interval.
(b) Find the total distance travelled in the 9 seconds.
- \(\int_{0}^{9}|v|\,dt=28.8\)
- The velocity changes sign at both times, so use the absolute value.
(c) Find the displacement from the starting point after 9 seconds.
- \(\int_{0}^{9}v\,dt=−7.97\)
jmaths.xyz · AI HL · AHL 2.9 Kinematics: displacement, velocity and acceleration