Kinematics on a line Problem set B One question at a time

The displacement \(s\) metres of an object from a fixed point at time \(t\) seconds is given. Find expressions for the velocity and acceleration.

1. (a) \(s = 5\sin 3t\)

   • \(v = \dfrac{ds}{dt} = 15\cos 3t\)
   • \(a = \dfrac{dv}{dt} = −45\sin 3t\)
   • Chain rule each time.

2. (b) \(s = 2e^{-4t}\)

   • \(v = \dfrac{ds}{dt} = −8e^{-4t}\)
   • \(a = \dfrac{dv}{dt} = 32e^{-4t}\)
   • Chain rule each time.

3. (c) \(s = (5t - 2)^{3}\)

   • \(v = \dfrac{ds}{dt} = 15 (5t - 2)^{2}\)
   • \(a = \dfrac{dv}{dt} = 750t - 300\)
   • Chain rule each time.

A particle moves on a line. Its velocity \(v\) is given in terms of its displacement \(s\). Find an expression for the acceleration in terms of \(s\).

1. (a) \(v = 2s + 5\)

   • \(\dfrac{dv}{ds} = 2\)
   • \(a = v\dfrac{dv}{ds} = (2s + 5)(2)\)
   • \(a = 4s + 10\)

2. (b) \(v = s^{2} - 6\)

   • \(\dfrac{dv}{ds} = 2s\)
   • \(a = v\dfrac{dv}{ds} = (s^{2} - 6)(2s)\)
   • \(a = 2s^{3} - 12s\)

3. (c) \(v = e^{2s}\)

   • \(\dfrac{dv}{ds} = 2e^{2s}\)
   • \(a = v\dfrac{dv}{ds} = (e^{2s})(2e^{2s})\)
   • \(a = 2e^{4s}\)

Use your calculator to evaluate each definite integral. Give each answer to 3 significant figures. Time is in seconds.

1. (a) \(\displaystyle\int_0^4 8\cos(0.5t)\,dt\) (change in velocity, \(a = 8\cos(0.5t)\))

   • \(\text{change in } v = \displaystyle\int_0^4 8\cos(0.5t)\,dt\)
   • From the calculator, in radians: 14.5 m s⁻¹

2. (b) \(\displaystyle\int_0^{1.5} 6te^{-2t}\,dt\) (change in displacement, \(v = 6te^{-2t}\))

   • \(\text{change in } s = \displaystyle\int_0^{1.5} 6te^{-2t}\,dt\)
   • From the calculator, in radians: 1.20 m

A mass hangs on a spring and moves up and down. Its displacement from the rest position is \(s = 0.12\cos 8t\) metres at time \(t\) seconds.

1. (a) Find the velocity when \(t = 0.25\).

   • \(v = −0.96\sin 8t\)
   • From the calculator, in radians: \(v(0.25)\) = −0.873 m s⁻¹

2. (b) Find the acceleration when \(t = 0.25\).

   • \(a = −7.68\cos 8t\)
   • From the calculator, in radians: \(a(0.25)\) = 3.20 m s⁻²

A boat drifts towards a pier with the engine in reverse. Its velocity is \(v = 18 - 2s\) m s⁻¹ when it is \(s\) metres from the point where the engine was reversed.

1. (a) Find an expression for the acceleration in terms of \(s\).

   • \(\dfrac{dv}{ds} = −2\)
   • \(a = v\dfrac{dv}{ds} = 4s - 36\)

2. (b) Find the acceleration when \(s = 5\). What does the sign tell you?

   • \(a = 4s - 36\) at \(s = 5\): −16 m s⁻²
   • Negative: the boat is slowing down.

A skater pushes off a wall and her acceleration is \(a = 7e^{-0.5t}\) m s⁻², with \(t\) in seconds.

1. (a) Find the change in velocity between \(t = 1\) and \(t = 5\).

   • \(\text{change in } v = \displaystyle\int_{1}^{5} 7e^{-0.5t}\,dt\)
   • From the calculator: 7.34 m s⁻¹

2. (b) The velocity is 8 m s⁻¹ when \(t = 1\). Find the velocity when \(t = 5\).

   • \(v(5) = v(1) + \text{change} = 8 + 7.34\)
   • 15.3 m s⁻¹

A sensor on a vibrating plate moves on a line with displacement \(s = 3\cos 4t + 2\) millimetres.

1. (a) Find an expression for \(a\) in terms of \(t\).

   • \(v = −12\sin 4t\)
   • \(a = −48\cos 4t\)

2. (b) Find the constant \(k\) such that \(a = k(s - 2)\).

   • \(s - 2 = 3\cos 4t\)
   • \(a = −48\cos 4t = (−16)(3\cos 4t)\)
   • \(k = −16\)

A bead slides on a wire. Its velocity satisfies \(v^2 = 25 - 4s^2\). Use \(a = \dfrac{d}{ds}\left(\tfrac12 v^2\right)\) to find an expression for the acceleration in terms of \(s\).

• \(a = \dfrac{d}{ds}\left(\tfrac12 v^2\right)\)
• \(\tfrac12 v^2 = \frac{25}{2} - 2s^{2}\)
• \(a = −4s\)