Kinematics on a line Problem set A Worked solutions

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) \(s = 4\cos 5t\)

   • \(v = \dfrac{ds}{dt} = −20\sin 5t\)
   • \(a = \dfrac{dv}{dt} = −100\cos 5t\)
   • Chain rule each time.

2. (b) \(s = 3e^{-2t}\)

   • \(v = \dfrac{ds}{dt} = −6e^{-2t}\)
   • \(a = \dfrac{dv}{dt} = 12e^{-2t}\)
   • Chain rule each time.

3. (c) \(s = (2t + 1)^{4}\)

   • \(v = \dfrac{ds}{dt} = 8 (2t + 1)^{3}\)
   • \(a = \dfrac{dv}{dt} = 48 (2t + 1)^{2}\)
   • 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) \(v = 3s + 2\)

   • \(\dfrac{dv}{ds} = 3\)
   • \(a = v\dfrac{dv}{ds} = (3s + 2)(3)\)
   • \(a = 9s + 6\)

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

   • \(\dfrac{dv}{ds} = −2s\)
   • \(a = v\dfrac{dv}{ds} = (4 - s^{2})(−2s)\)
   • \(a = 2s^{3} - 8s\)

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

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

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

1. (a) \(\displaystyle\int_1^3 5\sin 2t\,dt\) (change in velocity, \(a = 5\sin 2t\))

   • \(\text{change in } v = \displaystyle\int_1^3 5\sin 2t\,dt\)
   • From the calculator, in radians: −3.44 m s⁻¹

2. (b) \(\displaystyle\int_0^2 3te^{-t}\,dt\) (change in displacement, \(v = 3te^{-t}\))

   • \(\text{change in } s = \displaystyle\int_0^2 3te^{-t}\,dt\)
   • From the calculator, in radians: 1.78 m

A piston rod moves in a straight line. Its displacement from the centre of the cylinder is \(s = 0.04\sin 50t\) metres at time \(t\) seconds.

1. (a) Find the velocity when \(t = 0.2\).

   • \(v = 2\cos 50t\)
   • From the calculator, in radians: \(v(0.2)\) = −1.68 m s⁻¹

2. (b) Find the acceleration when \(t = 0.2\).

   • \(a = −100\sin 50t\)
   • From the calculator, in radians: \(a(0.2)\) = 54.4 m s⁻²

A cyclist brakes as she approaches a junction. Her velocity is \(v = 24 - 3s\) m s⁻¹ when she is \(s\) metres past the point where she starts braking.

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

   • \(\dfrac{dv}{ds} = −3\)
   • \(a = v\dfrac{dv}{ds} = 9s - 72\)

2. (b) Find the acceleration when \(s = 4\). What does the sign tell you?

   • \(a = 9s - 72\) at \(s = 4\): −36 m s⁻²
   • Negative: she is slowing down.

A rocket sled's acceleration is \(a = 9e^{-0.3t}\) m s⁻², with \(t\) in seconds.

1. (a) Find the change in velocity between \(t = 2\) and \(t = 6\).

   • \(\text{change in } v = \displaystyle\int_{2}^{6} 9e^{-0.3t}\,dt\)
   • From the calculator: 11.5 m s⁻¹

2. (b) The velocity is 15 m s⁻¹ when \(t = 2\). Find the velocity when \(t = 6\).

   • \(v(6) = v(2) + \text{change} = 15 + 11.5\)
   • 26.5 m s⁻¹

A particle moves on a line with displacement \(s = 5\sin 2t + 1\) metres.

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

   • \(v = 10\cos 2t\)
   • \(a = −20\sin 2t\)

2. (b) Find the constant \(k\) such that \(a = k(s - 1)\).

   • \(s - 1 = 5\sin 2t\)
   • \(a = −20\sin 2t = (−4)(5\sin 2t)\)
   • \(k = −4\)

A particle moves on a line. Its velocity satisfies \(v^2 = 16 - s^2\). Use \(a = \dfrac{d}{ds}\left(\tfrac12 v^2\right)\) to find an expression for the acceleration in terms of \(s\).

• \(a = \dfrac{d}{ds}\left(\tfrac12 v^2\right)\)
• \(\tfrac12 v^2 = 8 - \frac{s^{2}}{2}\)
• \(a = -s\)