AI HL · AHL 4
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Motion on a line: direction changes, distance and speed
AHL 4.22
Screenshot and video to come
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A cyclist rides along a straight road. Her velocity is \(v = 2t^{2} − 12t + 16\) m s⁻¹ at time \(t\) seconds, for \(t \ge 0\), where positive means away from home. Find the times when she is at rest, and say whether she changes direction each time.
Have a go first. Then open the working one line at a time.
Polynomial solver: \(2t^{2} − 12t + 16 = 0\), \(t = 2\) or \(t = 4\)
At rest means the velocity is zero.
\(v(1) = 6\), \(v(3) = −2\), \(v(5) = 6\)
Put one time before, between and after the two zeros into v.
Positive, then negative, then positive
v changes sign at both t = 2 and t = 4, so she turns round at both times: away from home, then back towards it, then away again.
The cyclist's velocity is \(v = 2t^{2} − 12t + 16\) m s⁻¹, for \(t \ge 0\), as before. Find her displacement from the start and the total distance she rides in the first 5 seconds.
Have a go first. Then open the working one line at a time.
From the calculator: \(\int_0^5 (2t^{2} − 12t + 16)\,dt = 13.3\) m
The change in position. The negative part cancels part of the positive.
From the calculator: \(\int_0^5 \left|2t^{2} − 12t + 16\right|\,dt = 18.7\) m
Distance is the integral of |v|: every part counts as positive.
\(13.3 + |−2.67| + 2.67 = 18.7\)
Split at the times she is at rest, t = 2 and t = 4, to see the three stages: forwards, back, forwards.
A drone flies along a straight line. Its velocity is \(v = 10\sin(0.6t) − 4\) m s⁻¹ for \(0 \le t \le 6\). Find the greatest velocity, and the greatest speed.
Have a go first. Then open the working one line at a time.
\(a = \dfrac{dv}{dt} = 6\cos(0.6t) = 0\), \(t = 2.62\), \(v = 6\)
The greatest velocity is where the acceleration is zero (check it is a maximum), or at an end.
\(v(0) = −4\), \(v(6) = −8.43\)
Always check both ends of the interval.
Greatest velocity 6 m s⁻¹, least velocity −8.43 m s⁻¹
Compare the three values.
Greatest speed \(|−8.43| = 8.43\) m s⁻¹
The greatest speed is the biggest |v|. It is not the greatest velocity here: the drone is faster backwards at t = 6.
A bead slides along a straight wire. Its displacement from the centre is \(s = t^{3} − 12t^{2} + 36t\) cm at time \(t\) seconds. Decide whether the bead is speeding up or slowing down when \(t = 1\), \(t = 3\) and \(t = 5\).
Have a go first. Then open the working one line at a time.
\(v = 3t^{2} − 24t + 36\), \(a = 6t − 24\)
Differentiate s, then differentiate v.
\(v = 15\), \(a = −18\): opposite signs, slowing down
Moving forwards but accelerating backwards.
\(v = −9\), \(a = −6\): same sign, speeding up
Both negative: the bead is moving backwards and the acceleration is also backwards.
\(v = −9\), \(a = 6\): opposite signs, slowing down
Moving backwards, accelerating forwards.
A tram moves along a straight track. The graph shows its velocity \(v\) in m s⁻¹ at time \(t\) seconds, where positive means towards the depot. (a) Find the acceleration during the first 3 seconds. (b) Find the time when the tram changes direction. (c) Find the total distance it travels in the 15 seconds.
Have a go first. Then open the working one line at a time.
\(a = \dfrac{6 − 0}{3 − 0} = 2\) m s⁻²
Acceleration is the gradient of the velocity-time graph.
\(v = 0\) at \(t = 12\)
The line from (9, 6) to (13, −2) falls 2 for every 1 second, so it reaches zero 3 seconds after t = 9. The velocity changes sign there.
\(\tfrac12(3)(6) + (6)(6) + \tfrac12(3)(6) = 9 + 36 + 9 = 54\) m
The three pieces from t = 0 to t = 12: a triangle, a rectangle and a triangle.
\(\tfrac12(1)(2) + \tfrac12(2)(2) = 1 + 2 = 3\) m
From t = 12 to t = 15 the tram is going backwards.
\(54 + 3 = 57\) m
Add both as positive numbers. The displacement would be 54 − 3 = 51 m.
A runner jogs along a straight path. The graph shows her displacement \(s\) metres from a gate at time \(t\) seconds. (a) Find the times when she is at rest. (b) Find the total distance she runs in the first 6 seconds.
Have a go first. Then open the working one line at a time.
The gradient of the curve is zero at \(t = 2\) and \(t = 4\)
Velocity is the gradient of the displacement-time graph. It is zero at the top and the bottom of the bends.
\(s(0) = 0\), \(s(2) = 20\), \(s(4) = 16\), \(s(6) = 36\)
Read the heights of the curve from the grid at t = 0, 2, 4 and 6.
\(20 + 4 + 20 = 44\) m
Forwards 20, backwards 4, forwards 20. The distances are added, not subtracted.
A particle moves on a straight line, starting at the origin. Its velocity is \(v = 3t^{2} − 14t + 8\) m s⁻¹ at time \(t\) seconds. Find the first time after the start when it is back at the origin.
Have a go first. Then open the working one line at a time.
\(s = \int v\,dt = t^{3} − 7t^{2} + 8t\) (no constant: \(s = 0\) when \(t = 0\))
Integrate the velocity. The start is at the origin, so the constant is zero.
Polynomial solver: \(t^{3} − 7t^{2} + 8t = 0\), \(t = 0\), \(t = 1.44\), \(t = 5.56\)
t = 0 is the start. The next solution is the first return.
\(t = 1.44\) s
It moves away in the positive direction first (v is positive at the start) and comes back through the origin at this time.