Find position by integrating velocity

5. A t time \(t\) seconds ( \(t \geqslant 0\) ), a particle \(P\) has velocity \(\mathbf { v m ~ s } ^ { - 1 }\), where $$\mathbf { v } = \left( 3 t ^ { 2 } - 4 \right) \mathbf { i } + ( 2 t - 4 ) \mathbf { j }$$ When \(t = 0 , P\) is at the fixed point \(O\).

1. Find the acceleration of \(P\) at the instant when \(t = 0\)
2. Find the exact speed of \(P\) at the instant when \(P\) is moving in the direction of the vector \(( 11 \mathbf { i } + \mathbf { j } )\) for the second time.
3. Show that \(P\) never returns to \(O\). \includegraphics[max width=\textwidth, alt={}, center]{c16c17b6-2c24-4939-b3b5-63cd63646b76-14_2658_1938_107_123} \includegraphics[max width=\textwidth, alt={}, center]{c16c17b6-2c24-4939-b3b5-63cd63646b76-15_149_140_2604_1818}

1. \hspace{0pt} [In this question, the perpendicular unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are in a horizontal plane.]

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. A particle \(P\) is moving on a smooth horizontal plane.
At time \(t\) seconds \(( t \geqslant 0 )\), the position vector of \(P\), relative to a fixed point \(O\), is \(\mathbf { r }\) metres and the velocity of \(P\) is \(\mathbf { v } \mathrm { ms } ^ { - 1 }\) where $$\mathbf { v } = \left( 4 t ^ { 2 } - 5 t \right) \mathbf { i } + ( - 10 t - 12 ) \mathbf { j }$$ When \(t = 0 , \mathbf { r } = 2 \mathbf { i } + 6 \mathbf { j }\)

1. Find \(\mathbf { r }\) when \(t = 2\) When \(t = T\) particle \(P\) is moving in the direction of the vector \(\mathbf { i } - 2 \mathbf { j }\)
2. Find the value of \(T\)
3. Find the exact magnitude of the acceleration of \(P\) when \(t = 2.5\)

4. At time \(t\) seconds \(( 0 \leqslant t < 5 )\), a particle \(P\) has velocity \(\mathbf { v m s } ^ { - 1 }\), where $$\mathbf { v } = ( \sqrt { 5 - t } ) \mathbf { i } + \left( t ^ { 2 } + 2 t - 3 \right) \mathbf { j }$$ When \(t = \lambda\), particle \(P\) is moving in a direction parallel to the vector \(\mathbf { i }\).

1. Find the acceleration of \(P\) when \(t = \lambda\) The position vector of \(P\) is measured relative to the fixed point \(O\) When \(t = 1\), the position vector of \(P\) is \(( - 2 \mathbf { i } + \mathbf { j } ) \mathrm { m }\). Given that \(1 \leqslant T < 5\)
2. find, in terms of \(T\), the position vector of \(P\) when \(t = T\)

8 A toy boat moves in a horizontal plane with position vector \(\mathbf { r } = x \mathbf { i } + y \mathbf { j }\), where \(\mathbf { i }\) and \(\mathbf { j }\) are the standard unit vectors east and north respectively. The origin of the position vectors is at O . The displacements \(x\) and \(y\) are in metres. First consider only the motion of the boat parallel to the \(x\)-axis. For this motion $$x = 8 t - 2 t ^ { 2 }$$ The velocity of the boat in the \(x\)-direction is \(v _ { x } \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Find an expression in terms of \(t\) for \(v _ { x }\) and determine when the boat instantaneously has zero speed in the \(x\)-direction. Now consider only the motion of the boat parallel to the \(y\)-axis. For this motion $$v _ { y } = ( t - 2 ) ( 3 t - 2 )$$ where \(v _ { y } \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the velocity of the boat in the \(y\)-direction at time \(t\) seconds.
2. Given that \(y = 3\) when \(t = 1\), use integration to show that \(y = t ^ { 3 } - 4 t ^ { 2 } + 4 t + 2\). The position vector of the boat is given in terms of \(t\) by \(\mathbf { r } = \left( 8 t - 2 t ^ { 2 } \right) \mathbf { i } + \left( t ^ { 3 } - 4 t ^ { 2 } + 4 t + 2 \right) \mathbf { j }\).
3. Find the time(s) when the boat is due north of O and also the distance of the boat from O at any such times.
4. Find the time(s) when the boat is instantaneously at rest. Find the distance of the boat from O at any such times.
5. Plot a graph of the path of the boat for \(0 \leqslant t \leqslant 2\).

1. \hspace{0pt} [In this question position vectors are given relative to a fixed origin \(O\) ]

At time \(t\) seconds, where \(t \geqslant 0\), a particle, \(P\), moves so that its velocity \(\mathbf { v ~ m ~ s } ^ { - 1 }\) is given by $$\mathbf { v } = 6 t \mathbf { i } - 5 t ^ { \frac { 3 } { 2 } } \mathbf { j }$$ When \(t = 0\), the position vector of \(P\) is \(( - 20 \mathbf { i } + 20 \mathbf { j } ) \mathrm { m }\).

1. Find the acceleration of \(P\) when \(t = 4\)
2. Find the position vector of \(P\) when \(t = 4\)

1. \hspace{0pt} [In this question, position vectors are given relative to a fixed origin.] At time \(t\) seconds, where \(t > 0\), a particle \(P\) has velocity \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\) where

$$\mathbf { v } = 3 t ^ { 2 } \mathbf { i } - 6 t ^ { \frac { 1 } { 2 } } \mathbf { j }$$

1. Find the speed of \(P\) at time \(t = 2\) seconds.
2. Find an expression, in terms of \(t , \mathbf { i }\) and \(\mathbf { j }\), for the acceleration of \(P\) at time \(t\) seconds, where \(t > 0\) At time \(t = 4\) seconds, the position vector of \(P\) is ( \(\mathbf { i } - 4 \mathbf { j }\) ) m.
3. Find the position vector of \(P\) at time \(t = 1\) second.

1. At time \(t\) seconds, a particle \(P\) has velocity \(\mathbf { v } \mathrm { ms } ^ { - 1 }\), where

$$\mathbf { v } = 3 t ^ { \frac { 1 } { 2 } } \mathbf { i } - 2 t \mathbf { j } \quad t > 0$$

1. Find the acceleration of \(P\) at time \(t\) seconds, where \(t > 0\)
2. Find the value of \(t\) at the instant when \(P\) is moving in the direction of \(\mathbf { i } - \mathbf { j }\) At time \(t\) seconds, where \(t > 0\), the position vector of \(P\), relative to a fixed origin \(O\), is \(\mathbf { r }\) metres. When \(t = 1 , \mathbf { r } = - \mathbf { j }\)
3. Find an expression for \(\mathbf { r }\) in terms of \(t\).
4. Find the exact distance of \(P\) from \(O\) at the instant when \(P\) is moving with speed \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\)

9 In this question, the vectors \(\mathbf { i }\) and \(\mathbf { j }\) are directed east and north respectively.
The velocity \(\mathbf { v } \mathrm { ms } ^ { - 1 }\) of a particle at time \(t\) s is given by \(\mathbf { v } = k t ^ { 2 } \mathbf { i } + 6 t\), where \(k\) is a positive constant. The magnitude of the acceleration when \(t = 2\) is \(10 \mathrm {~ms} ^ { - 2 }\).

1. Calculate the value of \(k\). The particle is at the origin when \(t = 0\).
2. Determine an expression for the position vector of the particle at time \(t\).
3. Determine the time when the particle is directly north-east of the origin.

3. A particle \(P\) moves in a horizontal plane such that, at time \(t\) seconds, its velocity is \(\mathbf { v } \mathrm { ms } ^ { - 1 }\), where \(\mathbf { v } = 2 t \mathbf { i } - t ^ { \frac { 1 } { 2 } } \mathbf { j }\). When \(t = 0 , P\) is at the point with position vector \(- 10 \mathbf { i } + \mathbf { j }\) relative to a fixed origin \(O\).

1. Find the position vector \(\mathbf { r }\) of \(P\) at time \(t\) seconds.
2. Find the distance \(O P\) when \(t = 4\).

A particle moves in a straight line with acceleration \(a\) m s\(^{-2}\), at time \(t\) seconds, where $$a = 10 - 6t$$ The particle's velocity, \(v\) m s\(^{-1}\), and displacement, \(r\) metres, are both initially zero. Show that $$r = t^2(5 - t)$$ Fully justify your answer. [4 marks]