Answer :
Let's work through this question step by step.
### (ii) Determine the velocity of the particle at [tex]\( t = 3 \)[/tex] seconds.
Given the velocity function:
[tex]\[ V(t) = 5t - t^2 + 2 \][/tex]
We need to find [tex]\( V \)[/tex] at [tex]\( t = 3 \)[/tex] seconds:
[tex]\[ \begin{aligned} V(3) &= 5(3) - (3)^2 + 2 \\ &= 15 - 9 + 2 \\ &= 8 \, \text{m/s} \\ \end{aligned} \][/tex]
So, the velocity of the particle at [tex]\( t = 3 \)[/tex] seconds is [tex]\( 8 \, \text{m/s} \)[/tex].
### (b) Find the time taken by the particle to attain maximum velocity and the distance it covered to attain the maximum velocity.
To find the time at which the particle attains maximum velocity, we need to find the critical points of the velocity function [tex]\( V(t) \)[/tex]. This is done by finding the derivative of [tex]\( V(t) \)[/tex] and setting it to zero.
Given:
[tex]\[ V(t) = 5t - t^2 + 2 \][/tex]
The derivative [tex]\( V'(t) \)[/tex] is:
[tex]\[ V'(t) = \frac{d}{dt}(5t - t^2 + 2) = 5 - 2t \][/tex]
Setting [tex]\( V'(t) = 0 \)[/tex] to find the critical points:
[tex]\[ \begin{aligned} 5 - 2t &= 0 \\ 2t &= 5 \\ t &= \frac{5}{2} \\ t &= 2.5 \, \text{seconds} \\ \end{aligned} \][/tex]
So, the time at which the particle attains maximum velocity is [tex]\( t = 2.5 \)[/tex] seconds.
Next, we need to find the distance covered by the particle to attain the maximum velocity. The distance [tex]\( S(t) \)[/tex] is the integral of the velocity function [tex]\( V(t) \)[/tex].
Given:
[tex]\[ V(t) = 5t - t^2 + 2 \][/tex]
The integral of [tex]\( V(t) \)[/tex] with respect to [tex]\( t \)[/tex] is:
[tex]\[ \begin{aligned} S(t) &= \int (5t - t^2 + 2) \, dt \\ &= \int 5t \, dt - \int t^2 \, dt + \int 2 \, dt \\ &= \frac{5t^2}{2} - \frac{t^3}{3} + 2t + C \\ \end{aligned} \][/tex]
To find the distance covered at the time of maximum velocity ([tex]\( t = 2.5 \)[/tex]), we set the constant [tex]\( C \)[/tex] to zero (assuming initial distance at [tex]\( t = 0 \)[/tex] is zero for simplicity):
[tex]\[ S(t) = \frac{5t^2}{2} - \frac{t^3}{3} + 2t \][/tex]
Substituting [tex]\( t = 2.5 \)[/tex]:
[tex]\[ \begin{aligned} S(2.5) &= \frac{5 \cdot (2.5)^2}{2} - \frac{(2.5)^3}{3} + 2 \cdot 2.5 \\ &= \frac{5 \cdot 6.25}{2} - \frac{15.625}{3} + 5 \\ &= \frac{31.25}{2} - 5.2083 + 5 \\ &= 15.625 - 5.2083 + 5 \\ &= 15.4167 \, \text{meters} \\ \end{aligned} \][/tex]
Thus, the distance covered by the particle to attain maximum velocity is approximately [tex]\( 15.4167 \, \text{meters} \)[/tex].
### (ii) Determine the velocity of the particle at [tex]\( t = 3 \)[/tex] seconds.
Given the velocity function:
[tex]\[ V(t) = 5t - t^2 + 2 \][/tex]
We need to find [tex]\( V \)[/tex] at [tex]\( t = 3 \)[/tex] seconds:
[tex]\[ \begin{aligned} V(3) &= 5(3) - (3)^2 + 2 \\ &= 15 - 9 + 2 \\ &= 8 \, \text{m/s} \\ \end{aligned} \][/tex]
So, the velocity of the particle at [tex]\( t = 3 \)[/tex] seconds is [tex]\( 8 \, \text{m/s} \)[/tex].
### (b) Find the time taken by the particle to attain maximum velocity and the distance it covered to attain the maximum velocity.
To find the time at which the particle attains maximum velocity, we need to find the critical points of the velocity function [tex]\( V(t) \)[/tex]. This is done by finding the derivative of [tex]\( V(t) \)[/tex] and setting it to zero.
Given:
[tex]\[ V(t) = 5t - t^2 + 2 \][/tex]
The derivative [tex]\( V'(t) \)[/tex] is:
[tex]\[ V'(t) = \frac{d}{dt}(5t - t^2 + 2) = 5 - 2t \][/tex]
Setting [tex]\( V'(t) = 0 \)[/tex] to find the critical points:
[tex]\[ \begin{aligned} 5 - 2t &= 0 \\ 2t &= 5 \\ t &= \frac{5}{2} \\ t &= 2.5 \, \text{seconds} \\ \end{aligned} \][/tex]
So, the time at which the particle attains maximum velocity is [tex]\( t = 2.5 \)[/tex] seconds.
Next, we need to find the distance covered by the particle to attain the maximum velocity. The distance [tex]\( S(t) \)[/tex] is the integral of the velocity function [tex]\( V(t) \)[/tex].
Given:
[tex]\[ V(t) = 5t - t^2 + 2 \][/tex]
The integral of [tex]\( V(t) \)[/tex] with respect to [tex]\( t \)[/tex] is:
[tex]\[ \begin{aligned} S(t) &= \int (5t - t^2 + 2) \, dt \\ &= \int 5t \, dt - \int t^2 \, dt + \int 2 \, dt \\ &= \frac{5t^2}{2} - \frac{t^3}{3} + 2t + C \\ \end{aligned} \][/tex]
To find the distance covered at the time of maximum velocity ([tex]\( t = 2.5 \)[/tex]), we set the constant [tex]\( C \)[/tex] to zero (assuming initial distance at [tex]\( t = 0 \)[/tex] is zero for simplicity):
[tex]\[ S(t) = \frac{5t^2}{2} - \frac{t^3}{3} + 2t \][/tex]
Substituting [tex]\( t = 2.5 \)[/tex]:
[tex]\[ \begin{aligned} S(2.5) &= \frac{5 \cdot (2.5)^2}{2} - \frac{(2.5)^3}{3} + 2 \cdot 2.5 \\ &= \frac{5 \cdot 6.25}{2} - \frac{15.625}{3} + 5 \\ &= \frac{31.25}{2} - 5.2083 + 5 \\ &= 15.625 - 5.2083 + 5 \\ &= 15.4167 \, \text{meters} \\ \end{aligned} \][/tex]
Thus, the distance covered by the particle to attain maximum velocity is approximately [tex]\( 15.4167 \, \text{meters} \)[/tex].
