Calculus - Kinematics

A particle moves in a straight line such that its displacement $s$ (in meters) from a fixed point $O$ at time $t$ (in seconds) is given by $s(t) = t^3 - 9t^2 + 24t + 5$ for $t \ge 0$. Find the time(s) when the particle is at rest.

The particle is at rest when $v(t) = 0$. First, find the velocity function by differentiating $s(t)$: $v(t) = s'(t) = 3t^2 - 18t + 24$ Set $v(t) = 0$: $3t^2 - 18t + 24 = 0$ Divide by $3$: $t^2 - 6t + 8 = 0$ Factorize the quadratic: $(t - 2)(t - 4) = 0$ $t = 2$ or $t = 4$

To find when a particle is at rest, we calculate the first derivative of displacement to get velocity and solve for $t$ when velocity equals zero.

The acceleration of a moving object is given by $a(t) = 6t - 4$. At $t = 0$, the velocity of the object is $10 \text{ m s}^{-1}$. Find an expression for the velocity $v(t)$ and calculate the velocity at $t = 3$.

Integrate acceleration to find velocity: $v(t) = \int (6t - 4) dt = 3t^2 - 4t + C$ Use the initial condition $v(0) = 10$ to find $C$: $10 = 3(0)^2 - 4(0) + C \Rightarrow C = 10$ So, $v(t) = 3t^2 - 4t + 10$ At $t = 3$: $v(3) = 3(3)^2 - 4(3) + 10 = 27 - 12 + 10 = 25 \text{ m s}^{-1}$

Velocity is the integral of acceleration. We use the given initial velocity (the value of $v$ when $t=0$) to solve for the constant of integration.

The velocity of a particle is given by $v(t) = 2t + 5$. Calculate the displacement of the particle between $t = 1$ and $t = 4$.

The displacement is the definite integral of the velocity function: $\int_{1}^{4} (2t + 5) dt$ Find the antiderivative: $[t^2 + 5t]_{1}^{4}$ Evaluate at the limits: $(4^2 + 5(4)) - (1^2 + 5(1))$ $(16 + 20) - (1 + 5)$ $36 - 6 = 30$ $\begin{array}{r} 36 \\ - 6 \\ \hline 30 \end{array}$ The displacement is $30 \text{ m}$.

To find displacement over a specific time interval, we calculate the definite integral of the velocity function.