Kinematics：練習問題

A particle moves along a straight line such that its velocity \(v \text{ m s}^{-1}\) at time \(t \text{ s}\) is given by \(v = 4t - t^2\) for the interval \(0 \le t \le 6\). Calculate the total distance travelled by the particle in this time interval.

The position vector \(\mathbf{r}\) of a particle at time \(t\) is given by \(\mathbf{r} = (3t^2)\mathbf{i} + (2t^3 - 6t)\mathbf{j}\). Calculate the magnitude of the velocity of the particle at the instant when the vertical component of its acceleration is \(24 \text{ m s}^{-2}\).

A particle moves in a straight line with constant acceleration of \(4.0 \text{ m s}^{-2}\). If its initial velocity is \(6.0 \text{ m s}^{-1}\), find the velocity of the particle after \(5.0 \text{ s}\).

A particle moves with constant acceleration along a straight line. If its initial velocity is \(u = 5 \text{ m s}^{-1}\) and its final velocity after \(4\) seconds is \(v = 17 \text{ m s}^{-1}\), determine the magnitude of the constant acceleration \(a\) in \(\text{m s}^{-2}\).

The velocity \(v \text{ m s}^{-1}\) of a particle moving in a straight line at time \(t \text{ s}\) is given by \(v = 6t - 2t^2\). Calculate the total distance travelled by the particle in the interval \(0 \le t \le 4\).

A stone is thrown vertically upwards from the ground with an initial speed of \(19.6 \text{ m s}^{-1}\). Taking the acceleration due to gravity as \(g = 9.8 \text{ m s}^{-2}\), calculate the time taken for the stone to reach its maximum height.

A particle P moves along a straight line. Its acceleration $a$ in $ \text{m s}^{-2}$ at time $t$ seconds, for $t \ge 0$, is given by $a = 6t - 8$.

When $t=0$, the displacement $s$ of the particle from the origin O is $4 \text{ m}$, and its velocity $v$ is $5 \text{ m s}^{-1}$.

(a) Find an expression for the velocity $v$ of the particle in terms of $t$.

(b) Find the displacement $s$ of the particle from O at time $t$.

(c) Find the total distance travelled by the particle in the interval $0 \le t \le 3$.