Step-by-Step Solution
Step 1
We have the given the following values:
The initial velocity of the particle is $u = 0\;{\rm{m}}/{\rm{s}}$.
The equation of acceleration is $a = \left( {2t – 6} \right)\;{\rm{m}}/{{\rm{s}}^2}$.
We are asked to calculate the particle’s velocity at t = 6 s and position at t = 11 s.
Step 2
The acceleration is given by:\[\begin{array}{c} a = \frac{{dv}}{{dt}}\\ dv = adt \end{array}\]
Substitute the value of a in the above equation:\[dv = \left( {2t – 6} \right)dt\]
Step 3
Integrate the above equation with the lower to upper limits of velocity (0 m/s to v) and time (0 s to t):\[\begin{array}{c} \int\limits_u^v {dv} = \int\limits_0^t {\left( {2t – 6} \right)dt} \\ \left[ v \right]_0^v = \left[ {{t^2} – 6t} \right]_0^t\\ \left( {v – 0} \right) = \left( {{t^2} – 6t – 0} \right)\\ v = {t^2} – 6t \end{array}\]……(1)
Substitute $t = 6\;{\rm{s}}$ in equation (1) to calculate the required velocity:\[\begin{array}{c} v = \left( {{{\left( 6 \right)}^2} – 6\left( 6 \right)} \right)\;{\rm{m}}/{\rm{s}}\\ v = 0\;{\rm{m}}/{\rm{s}} \end{array}\]
Step 4
The velocity of the particle is given by:\[\begin{array}{c} v = \frac{{ds}}{{dt}}\\ ds = vdt \end{array}\]
Here, s is the distance travelled by the particle.
Substitute the value of v from equation (1) in the above equation:\[ds = \left( {{t^2} – 6t} \right)dt\]
Step 5
Integrate the above equation with the lower to upper limits of distance travelled (0 m to s) and time (0 s to t):\[\begin{array}{c} \int\limits_0^s {ds} = \int\limits_0^t {\left( {{t^2} – 6t} \right)dt} \\ \left[ s \right]_0^s = \left[ {\frac{{{t^3}}}{3} – 3{t^2}} \right]_0^t\\ \left( {s – 0} \right) = \left( {\frac{{{t^3}}}{3} – 3{t^2} – 0} \right)\\ s = \frac{{{t^3}}}{3} – 3{t^2} \end{array}\]
Substitute $t = 11\;{\rm{s}}$ in above equation to calculate the required position of the particle:\[\begin{array}{c} s = \left( {\frac{{{{\left( {11} \right)}^3}}}{3} – 3{{\left( {11} \right)}^2}} \right)\;{\rm{m}}\\ = 80.67\;{\rm{m}} \end{array}\]