Distance from velocity function using calculus

6 A particle travels along a straight line. Its velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) after \(t\) seconds is given by $$v = t ^ { 3 } - 6 t ^ { 2 } + 8 t \text { for } 0 \leqslant t \leqslant 4$$ When \(t = 0\) the particle is at rest at the point \(P\).

1. Find the times (other than \(t = 0\) ) when the particle is at rest. Sketch the velocity-time graph for \(0 \leqslant t \leqslant 4\).
2. Find the acceleration of the particle when \(t = 2\).
3. Find an expression for the displacement of the particle from \(P\) after \(t\) seconds. Hence state its displacement from \(P\) when \(t = 2\) and find its average speed between \(t = 0\) and \(t = 2\).

A particle \(P\) moves in a straight line. The velocity \(v \text{ ms}^{-1}\) at time \(t\) s is given by $$v = 2t + 1 \quad \text{for } 0 \leqslant t \leqslant 5,$$ $$v = 36 - t^2 \quad \text{for } 5 \leqslant t \leqslant 7,$$ $$v = 2t - 27 \quad \text{for } 7 \leqslant t \leqslant 13.5.$$

1. Sketch the velocity-time graph for \(0 \leqslant t \leqslant 13.5\). [3]
2. Find the acceleration at the instant when \(t = 6\). [2]
3. Find the total distance travelled by \(P\) in the interval \(0 \leqslant t \leqslant 13.5\). [5]

\includegraphics{figure_7} A car \(P\) starts from rest and travels along a straight road for \(600\) s. The \((t, v)\) graph for the journey is shown in the diagram. This graph consists of three straight line segments. Find

1. the distance travelled by \(P\), [3]
2. the deceleration of \(P\) during the interval \(500 < t < 600\). [2]

Another car \(Q\) starts from rest at the same instant as \(P\) and travels in the same direction along the same road for \(600\) s. At time \(t\) s after starting the velocity of \(Q\) is \((600t^2 - t^3) \times 10^{-6}\) m s\(^{-1}\).

1. Find an expression in terms of \(t\) for the acceleration of \(Q\). [2]
2. Find how much less \(Q\)'s deceleration is than \(P\)'s when \(t = 550\). [2]
3. Show that \(Q\) has its maximum velocity when \(t = 400\). [2]
4. Find how much further \(Q\) has travelled than \(P\) when \(t = 400\). [6]

A motorcyclist starts from rest at a point \(O\) and travels in a straight line. His velocity after \(t\) seconds is \(v\) m s\(^{-1}\), for \(0 \leq t \leq T\), where \(v = 7.2t - 0.45t^2\). The motorcyclist's acceleration is zero when \(t = T\).

1. Find the value of \(T\). [4]
2. Show that \(v = 28.8\) when \(t = T\). [1]

For \(t \geq T\) the motorcyclist travels in the same direction as before, but with constant speed \(28.8\) m s\(^{-1}\).

1. Find the displacement of the motorcyclist from \(O\) when \(t = 31\). [6]

A cyclist travels along a straight road. Her velocity \(v\) m s\(^{-1}\), at time \(t\) seconds after starting from a point \(O\), is given by \(v = 2\) for \(0 \leq t \leq 10\), \(v = 0.03t^2 - 0.3t + 2\) for \(t \geq 10\).

1. Find the displacement of the cyclist from \(O\) when \(t = 10\). [1]
2. Show that, for \(t \geq 10\), the displacement of the cyclist from \(O\) is given by the expression \(0.01t^3 - 0.15t^2 + 2t + 5\). [4]
3. Find the time when the acceleration of the cyclist is \(0.6\) m s\(^{-2}\). Hence find the displacement of the cyclist from \(O\) when her acceleration is \(0.6\) m s\(^{-2}\). [5]

A particle starts from rest at a point \(A\) at time \(t = 0\), where \(t\) is in seconds. The particle moves in a straight line. For \(0 \leq t \leq 4\) the acceleration is \(1.8t\) m s\(^{-2}\), and for \(4 \leq t \leq 7\) the particle has constant acceleration \(7.2\) m s\(^{-2}\).

1. Find an expression for the velocity of the particle in terms of \(t\), valid for \(0 \leq t \leq 4\). [3]
2. Show that the displacement of the particle from \(A\) is \(19.2\) m when \(t = 4\). [4]
3. Find the displacement of the particle from \(A\) when \(t = 7\). [5]

A particle moves along the \(x\)-axis with velocity, \(v\) ms\(^{-1}\), at time \(t\) given by $$v = 24t - 6t^2.$$ The positive direction is in the sense of \(x\) increasing.

1. Find an expression for the acceleration of the particle at time \(t\). [2]
2. Find the times, \(t_1\) and \(t_2\), at which the particle has zero speed. [2]
3. Find the distance travelled between the times \(t_1\) and \(t_2\). [4]

The velocity, \(v\) ms\(^{-1}\), of a particle moving along a straight line is given by $$v = 3t^2 - 12t + 14,$$ where \(t\) is the time in seconds.

1. Find an expression for the acceleration of the particle at time \(t\). [2]
2. Find the displacement of the particle from its position when \(t = 1\) to its position when \(t = 3\). [4]
3. You are given that \(v\) is always positive. Explain how this tells you that the distance travelled by the particle between \(t = 1\) and \(t = 3\) has the same value as the displacement between these times. [2]

A box of emergency supplies is dropped to victims of a natural disaster from a stationary helicopter at a height of 1000 metres. The initial velocity of the box is zero. At time \(t\) s after being dropped, the acceleration, \(a\text{ m s}^{-2}\), of the box in the vertically downwards direction is modelled by $$a = 10 - t \text{ for } 0 \leqslant t \leqslant 10,$$ $$a = 0 \text{ for } t > 10.$$

1. Find an expression for the velocity, \(v\text{ m s}^{-1}\), of the box in the vertically downwards direction in terms of \(t\) for \(0 \leqslant t \leqslant 10\). Show that for \(t > 10\), \(v = 50\). [4]
2. Draw a sketch graph of \(v\) against \(t\) for \(0 \leqslant t \leqslant 20\). [3]
3. Show that the height, \(h\) m, of the box above the ground at time \(t\) s is given, for \(0 \leqslant t \leqslant 10\), by $$h = 1000 - 5t^2 + \frac{1}{6}t^3.$$ Find the height of the box when \(t = 10\). [4]
4. Find the value of \(t\) when the box hits the ground. [2]
5. Some of the supplies in the box are damaged when the box hits the ground. So measures are considered to reduce the speed with which the box hits the ground the next time one is dropped. Two different proposals are made. Carry out suitable calculations and then comment on each of them.
   1. The box should be dropped from a height of 500 m instead of 1000 m. [2]
   2. The box should be fitted with a parachute so that its acceleration is given by $$a = 10 - 2t \text{ for } 0 \leqslant t \leqslant 5,$$ $$a = 0 \text{ for } t > 5.$$ [3]