3.02f Non-uniform acceleration: using differentiation and integration

At time \(t\) seconds, \(t \geq 0\), a particle \(P\) has velocity \(\mathbf{v}\) m s\(^{-1}\), where $$\mathbf{v} = (27 - 3t^2)\mathbf{i} + (8 - t^3)\mathbf{j}$$ When \(t = 1\), the particle \(P\) is at the point with position vector \(\mathbf{r}\) m relative to a fixed origin \(O\), where \(\mathbf{r} = -5\mathbf{i} + 2\mathbf{j}\) Find

1. the magnitude of the acceleration of \(P\) at the instant when it is moving in the direction of the vector \(\mathbf{i}\), [5]
2. the position vector of \(P\) at the instant when \(t = 3\) [5]

\includegraphics{figure_2} At time \(t = 0\) a small package is projected from a point \(B\) which is \(2.4\) m above a point \(A\) on horizontal ground. The package is projected with speed \(23.75\) m s\(^{-1}\) at an angle \(α\) to the horizontal, where \(\tan α = \frac{4}{5}\). The package strikes the ground at point \(C\), as shown in Fig. 2. The package is modelled as a particle moving freely under gravity.

1. Find the time taken for the package to reach \(C\). [5]

A lorry moves along the line \(AC\), approaching \(A\) with constant speed 18 m s\(^{-1}\). At time \(t = 0\) the rear of the lorry passes \(A\) and the lorry starts to slow down. It comes to rest 7 seconds later. The acceleration, \(a\) m s\(^{-2}\) of the lorry at time \(t\) seconds is given by $$a = -\frac{1}{4}t^2, \quad 0 \leq t \leq 7.$$

1. Find the speed of the lorry at time \(t\) seconds. [3]

1. Hence show that \(T = 6\). [3]

1. Show that when the package reaches \(C\) it is just under 10 m behind the rear of the moving lorry. [5]

END

A particle \(P\) moves in a straight line so that, at time \(t\) seconds, its acceleration \(a\) m s\(^{-2}\) is given by $$a = \begin{cases} 4t - t^2, & 0 \leq t \leq 3, \\ \frac{27}{t^2}, & t > 3. \end{cases}$$ At \(t = 0\), \(P\) is at rest. Find the speed of \(P\) when

1. \(t = 3\), [3]

1. \(t = 6\). [5]

Figure 1 shows the path taken by a cyclist in travelling on a section of a road. When the cyclist comes to the point \(A\) on the top of a hill, she is travelling at 8 m s\(^{-1}\). She descends a vertical distance of 20 m to the bottom of the hill. The road then rises to the point \(B\) through a vertical distance of 12 m. When she reaches the point \(B\), her speed is 5 m s\(^{-1}\). The total mass of the cyclist and the cycle is 80 kg and the total distance along the road from \(A\) to \(B\) is 500 m. By modelling the resistance to the motion of the cyclist as of constant magnitude 20 N,

1. find the work done by the cyclist in moving from \(A\) to \(B\). [5]

At \(B\) the road is horizontal. Given that at \(B\) the cyclist is accelerating at 0.5 m s\(^{-2}\),

1. find the power generated by the cyclist at \(B\). [4]

A particle \(P\) moves on the \(x\)-axis. The acceleration of \(P\) at time \(t\) seconds is \((4t - 8)\) m s\(^{-2}\), measured in the direction of \(x\) increasing. The velocity of \(P\) at time \(t\) seconds is \(v\) m s\(^{-1}\). Given that \(v = 6\) when \(t = 0\), find

1. \(v\) in terms of \(t\), [4]

1. the distance between the two points where \(P\) is instantaneously at rest. [7]

A particle \(P\) moves on the \(x\)-axis. The acceleration of \(P\) at time \(t\) seconds is \((4t - 8)\) m s\(^{-2}\), measured in the direction of \(x\) increasing. The velocity of \(P\) at time \(t\) seconds is \(v\) m s\(^{-1}\). Given that \(v = 6\) when \(t = 0\), find

1. \(v\) in terms of \(t\), [4]
2. the distance between the two points where \(P\) is instantaneously at rest. [7]

A car of mass 800 kg is moving at a constant speed of 15 m s\(^{-1}\) down a straight road inclined at an angle \(\alpha\) to the horizontal, where \(\sin \alpha = \frac{3}{4}\). The resistance to motion from non-gravitational forces is modelled as a constant force of magnitude 900 N.

1. Find, in kW, the rate of working of the engine of the car. [4]

When the car is travelling down the road at 15 m s\(^{-1}\), the engine is switched off. The car comes to rest in time \(T\) seconds after the engine is switched off. The resistance to motion from non-gravitational forces is again modelled as a constant force of magnitude 900 N.

1. Find the value of \(T\). [4]

A particle \(P\) moves along the \(x\)-axis. At time \(t\) seconds the velocity of \(P\) is \(v \text{ ms}^{-1}\) in the positive \(x\)-direction, where \(v = 3t^2 - 4t + 3\). When \(t = 0\), \(P\) is at the origin \(O\). Find the distance of \(P\) from \(O\) when \(P\) is moving with minimum velocity. [8]

[In this question \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors in a horizontal and upward vertical direction respectively] A particle \(P\) is projected from a fixed point \(O\) on horizontal ground with velocity \(u(\mathbf{i} + c\mathbf{j}) \text{ ms}^{-1}\), where \(c\) and \(u\) are positive constants. The particle moves freely under gravity until it strikes the ground at \(A\), where it immediately comes to rest. Relative to \(O\), the position vector of a point on the path of \(P\) is \((x\mathbf{i} + y\mathbf{j})\) m.

1. Show that $$y = cx - \frac{4.9x^2}{u^2}.$$ [5]

Given that \(u = 7\), \(OA = R\) m and the maximum vertical height of \(P\) above the ground is \(H\) m,

1. using the result in part (a), or otherwise, find, in terms of \(c\),
   1. \(R\)
   2. \(H\).
   [6]

Given also that when \(P\) is at the point \(Q\), the velocity of \(P\) is at right angles to its initial velocity,

1. find, in terms of \(c\), the value of \(x\) at \(Q\). [6]

At time \(t\) seconds, a particle \(P\) has position vector \(r\) metres relative to a fixed origin \(O\), where $$\mathbf{r} = (t^2 + 2t)\mathbf{i} + (t - 2t^2)\mathbf{j}.$$ Show that the acceleration of \(P\) is constant and find its magnitude. [5]

\includegraphics{figure_2} At time \(t = 0\) a small package is projected from a point \(B\) which is 2.4 m above a point \(A\) on horizontal ground. The package is projected with speed 23.75 m s\(^{-1}\) at an angle \(\alpha\) to the horizontal, where \(\tan \alpha = \frac{4}{3}\). The package strikes the ground at the point \(C\), as shown in Fig. 2. The package is modelled as a particle moving freely under gravity.

1. Find the time taken for the package to reach \(C\). [5]

A lorry moves along the line \(AC\), approaching \(A\) with constant speed 18 m s\(^{-1}\). At time \(t = 0\) the rear of the lorry passes \(A\) and the lorry starts to slow down. It comes to rest \(T\) seconds later. The acceleration, \(a\) m s\(^{-2}\) of the lorry at time \(t\) seconds is given by $$a = -\frac{1}{4}t^2, \quad 0 \leq t \leq T.$$

1. Find the speed of the lorry at time \(t\) seconds. [3]
2. Hence show that \(T = 6\). [3]
3. Show that when the package reaches \(C\) it is just under 10 m behind the rear of the moving lorry. [5]

END

The velocity \(v\) m s\(^{-1}\) of a particle \(P\) at time \(t\) seconds is given by $$\mathbf{v} = (3t - 2)\mathbf{i} - 5t\mathbf{j}.$$

1. Show that the acceleration of \(P\) is constant. [2]

At \(t = 0\), the position vector of \(P\) relative to a fixed origin O is \(3\mathbf{i}\) m.

1. Find the distance of \(P\) from O when \(t = 2\). [6]

A particle \(P\) moves in a straight line so that, at time \(t\) seconds, its acceleration \(a\) m s\(^{-2}\) is given by $$a = \begin{cases} 4t - t^2, & 0 \leq t \leq 3, \\ \frac{27}{t^2}, & t > 3. \end{cases}$$ At \(t = 0\), \(P\) is at rest. Find the speed of \(P\) when

1. \(t = 3\), [3]
2. \(t = 6\). [5]

A particle \(P\) moves on the \(x\)-axis. At time \(t\) seconds the velocity of \(P\) is \(v\) m s\(^{-1}\) in the direction of \(x\) increasing, where \(v = 6t - 2t^2\). When \(t = 0\), \(P\) is at the origin \(O\). Find the distance of \(P\) from \(O\) when \(P\) comes to instantaneous rest after leaving \(O\). [5]

A particle \(P\) moves on the \(x\)-axis. At time \(t\) seconds, its acceleration is \((5 - 2t)\) m s\(^{-2}\), measured in the direction of \(x\) increasing. When \(t = 0\), its velocity is 6 m s\(^{-1}\) measured in the direction of \(x\) increasing. Find the time when \(P\) is instantaneously at rest in the subsequent motion. [6]

A particle \(P\) moves on the \(x\)-axis. The acceleration of \(P\) at time \(t\) seconds, \(t \geq 0\), is \((3t + 5)\) m s\(^{-2}\) in the positive \(x\)-direction. When \(t = 0\), the velocity of \(P\) is 2 m s\(^{-1}\) in the positive \(x\)-direction. When \(t = T\), the velocity of \(P\) is 6 m s\(^{-1}\) in the positive \(x\)-direction. Find the value of \(T\). (6)

A particle \(P\) moves on the \(x\)-axis. The acceleration of \(P\) at time \(t\) seconds is \((t - 4)\) m s\(^{-2}\) in the positive \(x\)-direction. The velocity of \(P\) at time \(t\) seconds is \(v\) m s\(^{-1}\). When \(t = 0\), \(v = 6\). Find

1. \(v\) in terms of \(t\), [4]
2. the values of \(t\) when \(P\) is instantaneously at rest, [3]
3. the distance between the two points at which \(P\) is instantaneously at rest. [4]

A particle \(P\) moves along a straight line in such a way that at time \(t\) seconds its velocity \(v\) m s\(^{-1}\) is given by $$v = \frac{1}{2}t^2 - 3t + 4$$ Find

1. the times when \(P\) is at rest, [4]
2. the total distance travelled by \(P\) between \(t = 0\) and \(t = 4\). [5]

A particle \(P\) of mass 2.5 kg moves along the positive \(x\)-axis. It moves away from a fixed origin \(O\), under the action of a force directed away from \(O\). When \(OP = x\) metres the magnitude of the force is \(2e^{-0.1x}\) newtons and the speed of \(P\) is \(v\) m s\(^{-1}\). When \(x = 0\), \(v = 2\). Find

1. \(v^2\) in terms of \(x\), [6]
2. the value of \(x\) when \(v = 4\). [3]
3. Give a reason why the speed of \(P\) does not exceed \(\sqrt{20}\) m s\(^{-1}\). [1]

A toy car of mass \(0.2\) kg is travelling in a straight line on a horizontal floor. The car is modelled as a particle. At time \(t = 0\) the car passes through a fixed point \(O\). After \(t\) seconds the speed of the car is \(v \text{ m s}^{-1}\) and the car is at a point \(P\) with \(OP = x\) metres. The resultant force on the car is modelled as \(\frac{1}{5}x(4 - 3x)\) N in the direction \(OP\). The car comes to instantaneous rest when \(x = 6\). Find

1. an expression for \(v^2\) in terms of \(x\), [7]
2. the initial speed of the car. [2]

A particle moving in a straight line starts from rest at the point \(O\) at time \(t = 0\). At time \(t\) seconds, the velocity \(v\) m s\(^{-1}\) of the particle is given by $$v = 3t(t - 4), \quad 0 \leq t \leq 5,$$ $$v = 75t^{-1}, \quad 5 \leq t \leq 10.$$

1. Sketch a velocity-time graph for the particle for \(0 \leq t \leq 10\). [3]
2. Find the set of values of \(t\) for which the acceleration of the particle is positive. [2]
3. Show that the total distance travelled by the particle in the interval \(0 \leq t \leq 5\) is \(39\) m. [3]
4. Find, to \(3\) significant figures, the value of \(t\) at which the particle returns to \(O\). [5]

A particle \(P\) is moving along the positive \(x\)-axis. At time \(t = 0\), \(P\) is at the origin \(O\). At time \(t\) seconds, \(P\) is \(x\) metres from \(O\) and has velocity \(v = 2e^{-t}\) m s\(^{-1}\) in the direction of \(x\) increasing.

1. Find the acceleration of \(P\) in terms of \(x\). [3]
2. Find \(x\) in terms of \(t\). [6]

\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]