Question

The equation for the distance a particle travels in a straight line from a point is $s = 45t + 11t - {t^3}$ . Find the velocity expression in terms of t, the acceleration expression in terms of t, and the velocity and acceleration when t= 3 seconds.

Answer 1

Hint: We will see the mathematical definition and relation between position, velocity and acceleration vector. The velocity vector is the derivative of the position vector. The acceleration vector is the derivative of the velocity vector.

Complete step by step answer:
We have given the position vector of a particle at time t,
$s = 45t + 11t - {t^3}$ (1)
We will find velocity vector of the particle at time t
We know that the velocity vector is the derivative of the position vector
We will find the derivative of equation 1
$ \Rightarrow v(t) = \dfrac{{ds}}{{dt}} = 45 + 11 - 3{t^2}$
$ \Rightarrow v(t) = \dfrac{{ds}}{{dt}} = 56 - 3{t^2}$ (2)
We will now find the acceleration vector of the particle at time t
We know that the acceleration vector is the derivative of the velocity vector
We will find the derivative of equation 2
$ \Rightarrow a(t) = \dfrac{{dv}}{{dt}} = - 6t$ (3)
We have to find the acceleration vector at time $t = 3$
We will substitute the value $t = 3$ in the equation 2 and 3 respectively
We have substitute $t = 3$ in equation 2
$ \Rightarrow v(t) = 56 - 3 \times 9m{s^{ - 1}}$
$ \Rightarrow v(t) = 29m{s^{ - 1}}$
We have substitute $t = 3$ in equation 3
$ \Rightarrow a(t) = - 6 \times 3m{s^{ - 2}}$
$ \Rightarrow a(t) = - 18m{s^{ - 2}}$
Hence, the velocity and acceleration of particle having position vector $s = 45t + 11t - {t^3}$ at time
 $t = 3$ is $ - 29m{s^{ - 1}}$ and $ - 18m{s^{ - 2}}$

Note:We have to take care of derivatives while solving these types of problems. To solve these types of questions, we have to be very good at the mathematical part too. We can also find the acceleration vector by doing a double derivative of the position vector.