Step 1

Part (a)

We have given the following values:

The given equation for the displacement is $s = \left( {2{t^3}} \right)\;{\rm{m}}$.

The time is $t = 2\;{\rm{s}}$.

We are asked to determine the velocity at $t = 2\;{\rm{s}}$.

The velocity v is given by:

\[v = \frac{{ds}}{{dt}}\]

Substitute the value of s in the above equation:

\[\begin{array}{l} v = \frac{{d\left( {2{t^3}} \right)}}{{dt}}\\ v = 6{t^2} \end{array}\]
 
Step 2

Substitute $t = 2\;{\rm{s}}$ in the above equation:

\[\begin{array}{c} {v_{t = 2\;{\rm{s}}}} = 6{\left( 2 \right)^2}\;{\rm{m}}/{\rm{s}}\\ = 24\;{\rm{m}}/{\rm{s}} \end{array}\]
 
Step 3

Part (b)

We have given the following values:

The velocity equation is $v = \left( {5s} \right)\;{\rm{m}}/{\rm{s}}$.

The distance travelled is $s = 1\;{\rm{m}}$.

We are asked to determine the acceleration at $s = 1\;{\rm{m}}$.

The given equation for the velocity is:

\[v = \left( {5s} \right)\;{\rm{m}}/{\rm{s}}\]

Here, s is the displacement in meter.

Differentiate the above equation with respect to displacement s:

\[\begin{array}{c} \frac{{dv}}{{ds}} = \frac{d}{{ds}}\left( {5s} \right)\\ \frac{{dv}}{{ds}} = 5\\ dv = 5ds \end{array}\]
 
Step 4

The acceleration a can be expressed as:

\[a = \frac{{vdv}}{{ds}}\]

Substitute the value of v and dv in the above equation:

\[\begin{array}{c} a = \frac{{\left( {5s} \right)\left( {5ds} \right)}}{{ds}}\\ a = 25s \end{array}\]
 
Step 5

Substitute $s = 1\;{\rm{m}}$ in the above equation to calculate the required acceleration:

\[\begin{array}{l} {a_{s = 1\;{\rm{m}}}} = 25\left( 1 \right)\;{\rm{m}}/{{\rm{s}}^2}\\ {a_{s = 1\;{\rm{m}}}} = 25\;{\rm{m}}/{{\rm{s}}^2} \end{array}\]
 
Step 6

Part (c)

We have given the following values:

The velocity equation is $v = \left( {4t + 5} \right)\;{\rm{m}}/{\rm{s}}$.

The time is $t = 2\;{\rm{s}}$.

We are asked to determine the acceleration at $t = 2\;{\rm{s}}$.

The given equation for the velocity is:

\[v = \left( {4t + 5} \right)\;{\rm{m}}/{\rm{s}}\]
 
Step 7

The acceleration is given by:

\[a = \frac{{dv}}{{dt}}\]

Substitute the given equation of v in the above equation:

\[\begin{array}{c} a = \frac{{d\left( {4t + 5} \right)}}{{dt}}\\ a = 4\;{\rm{m}}/{{\rm{s}}^2} \end{array}\]
 
Step 8

Substitute $t = 2\;{\rm{s}}$ in the above equation:

\[{a_{t = 2\;{\rm{s}}}} = 4\;{\rm{m}}/{{\rm{s}}^2}\]
 
Step 9

Part (d)

We have given the following values:

The acceleration is $a = 2\;{\rm{m}}/{{\rm{s}}^{\rm{2}}}$.

The time is $t = 2\;{\rm{s}}$.

The initial velocity at time $t = 0\;{\rm{s}}$ is $u = 0\;{\rm{m}}/{\rm{s}}$.

We are asked to determine the velocity at $t = 2\;{\rm{s}}$.

 
Step 10

Apply the first equation of motion:

\[v = u + at\]

Substitute the value of given parameters in above equation:

\[\begin{array}{c} v = \left( {0\;{\rm{m}}/{\rm{s}}} \right) + \left( {2\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {2\;{\rm{s}}} \right)\\ = 4\;{\rm{m}}/{\rm{s}} \end{array}\]
 
Step 11

Part (e)

We have given the following values:

The acceleration is $a = 2\;{\rm{m}}/{{\rm{s}}^2}$.

The distance travelled is $s = 4\;{\rm{m}}$.

The initial velocity at ${s_o} = 0\;{\rm{m}}$ is ${v_o} = 3\;{\rm{m}}/{\rm{s}}$.

We are asked to determine the velocity at $s = 4\;{\rm{m}}$.

 
Step 12

The required velocity v can be calculated by using the equation:

\[{v^2} = v_o^2 + 2a\left( {s - {s_o}} \right)\]

Substitute the value of given parameters in the above equation:

\[\begin{array}{c} {v^2} = {\left( {3\;{\rm{m}}/{\rm{s}}} \right)^2} + 2\left( {2\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {\left( {4\;{\rm{m}}} \right) - \left( {0\;{\rm{m}}} \right)} \right)\\ v = 5\,\;{\rm{m}}/{\rm{s}} \end{array}\]
 
Step 13

Part (f)

We have given the following values:

The acceleration is $a = \left( s \right)\;{\rm{m}}/{{\rm{s}}^2}$.

The distance travelled is $s = 5\;{\rm{m}}$.

The initial velocity at ${s_o} = 0\;{\rm{m}}$ is $v = 0\;{\rm{m}}/{\rm{s}}$.

We are asked to determine the velocity at $s = 5\;{\rm{m}}$.

 
Step 14

The relation between the acceleration, distance travelled, and velocity is given by:

\[a \times ds = v \times dv\]

Substitute the value of a in the above equation:

\[\left( s \right) \times ds = v \times dv\]
 
Step 15

Integrate the above equation with lower and upper limits of distance (4 m to 5 m) and velocity (0 m/s to v):

\[\begin{array}{c} \int\limits_{4\;{\rm{m}}}^{5\;{\rm{m}}} {s \times ds} = \int\limits_{0\;{\rm{m}}/{\rm{s}}}^v {v \times dv} \\ \left[ {\frac{{{s^2}}}{2}} \right]_{4\;{\rm{m}}}^{5\;{\rm{m}}} = \left[ {\frac{{{v^2}}}{2}} \right]_{0\;{\rm{m/s}}}^v\\ \left( {\frac{{{{\left( 5 \right)}^2}}}{2} - \frac{{{{\left( 4 \right)}^2}}}{2}} \right) = \left( {\frac{{{v^2}}}{2} - 0} \right)\\ v = 3\;{\rm{m}}/{\rm{s}} \end{array}\]
 
Step 16

Part (g)

We have given the following values:

The acceleration is $a = 4\;{\rm{m}}/{{\rm{s}}^2}$.

The time is $t = 3\;{\rm{s}}$.

The initial velocity is ${v_o} = 2\;{\rm{m}}/{\rm{s}}$.

The initial position is at ${s_o} = 2\;{\rm{m}}$.

We are asked to determine the position at $t = 3\;{\rm{s}}$.

 
Step 17

The required value of s can be calculated by using the equation:

\[s = {s_o} + {v_o}t + \frac{1}{2}a{t^2}\]

Substitute the value of given parameters in the above equation:

\[\begin{array}{c} s = \left( {2\;{\rm{m}}} \right) + \left( {2\;{\rm{m}}/{\rm{s}}} \right)\left( {3\;{\rm{s}}} \right) + \frac{1}{2}\left( {4\;{\rm{m}}/{{\rm{s}}^2}} \right){\left( {3\;{\rm{s}}} \right)^2}\\ = 26\,{\rm{m}} \end{array}\]
 
Step 18

Part (h)

We have given the following values:

The acceleration is $a = \left( {8{t^2}} \right)\;{\rm{m}}/{{\rm{s}}^2}$.

The time is $t = 1\;{\rm{s}}$.

The velocity at $t = 0\;{\rm{s}}$ is $v = 0\;{\rm{m}}/{\rm{s}}$.

We are asked to determine the velocity at $t = 1\;{\rm{s}}$.

 
Step 19

The acceleration can be given by the equation:

\[\begin{array}{c} a = \frac{{dv}}{{dt}}\\ dv = adt \end{array}\]

Substitute the value of a in the above equation:

\[dv = \left( {8{t^2}} \right)dt\]
 
Step 20

Integrate the above equation with the lower and upper limits of velocity (0 to v) and time (0 to 1 s):

\[\begin{array}{c} \int\limits_0^v {dv} = \int\limits_{0\;{\rm{s}}}^{1\;{\rm{s}}} {\left( {8{t^2}} \right)dt} \\ \left[ v \right]_0^v = \left[ {\frac{{8{t^3}}}{3}} \right]_{0\;{\rm{s}}}^{1\;{\rm{s}}}\\ \left( {v - 0} \right) = \left( {\frac{{8{{\left( 1 \right)}^3}}}{3} - \frac{{8{{\left( 0 \right)}^3}}}{3}} \right)\;{\rm{m}}/{\rm{s}}\\ v = 2.667\;{\rm{m}}/{\rm{s}} \end{array}\]
 
Step 21

Part (i)

We have given the following values:

The distance travelled is $s = \left( {3{t^2} + 2} \right)\;{\rm{m}}$.

The time is $t = 2\;{\rm{s}}$.

We are asked to determine the velocity at $t = 2\;{\rm{s}}$.

 
Step 22

The velocity is given by:

\[v = \frac{{ds}}{{dt}}\]

Substitute the value of s in the above equation:

\[\begin{array}{c} v = \frac{{d\left( {3{t^2} + 2} \right)}}{{dt}}\\ v = 6t \end{array}\]
 
Step 23

Substitute the value of t in the above equation:

\[\begin{array}{c} v = 6\left( 2 \right)\;{\rm{m}}/{\rm{s}}\\ = 12\;{\rm{m}}/{\rm{s}} \end{array}\]
 
Step 24

Part (j)

We have given the following value:

The distance between point A and B is ${s_{AB}} = 7\;{\rm{m}}$.

The distance between point B and C is ${s_{BC}} = 14\;{\rm{m}}$.

The time taken by the particle to move from point A to B is ${t_{AB}} = 4\;{\rm{s}}$.

The time taken by the particle to move from point B to C is ${t_{BC}} = 6\;{\rm{s}}$.

We are asked to determine the average velocity and average speed.

 
Step 25

The displacement between point A and C is given by:

\[{s_{AC}} = {s_{BC}} - {s_{AB}}\]

Substitute the value of ${s_{BC}}$ and ${s_{AB}}$ in the above equation:

\[\begin{array}{l} {s_{AC}} = \left( {14\;{\rm{m}}} \right) - \left( {7\;{\rm{m}}} \right)\\ {s_{AC}} = 7\;{\rm{m}} \end{array}\]
 
Step 26

The total distance travelled by the particle is given by:

\[s = {s_{AB}} + {s_{BC}}\]

Substitute the value of ${s_{BC}}$ and ${s_{AB}}$ in the above equation:

\[\begin{array}{l} s = \left( {14\;{\rm{m}}} \right) + \left( {7\;{\rm{m}}} \right)\\ s = 21\;{\rm{m}} \end{array}\]
 
Step 27

The total time taken by the particle to move from A to C is:

\[t = {t_{AB}} + {t_{BC}}\]

Substitute the value of ${t_{BC}}$ and ${t_{AB}}$ in the above equation:

\[\begin{array}{l} t = \left( {4\;{\rm{s}}} \right) + \left( {6\;{\rm{s}}} \right)\\ t = 10\;{\rm{s}} \end{array}\]
 
Step 28

The average velocity of the particle is given by:

\[{v_{avg}} = \frac{{{s_{AC}}}}{t}\]

Substitute the value of ${s_{AC}}$ and t in the above equation:

\[\begin{array}{c} {v_{avg}} = \frac{{\left( {7\;{\rm{m}}} \right)}}{{\left( {10\;{\rm{s}}} \right)}}\\ = 0.7\;{\rm{m}}/{\rm{s}} \end{array}\]
 
Step 29

The average speed of the particle is given by:

\[{\left( {{v_{sp}}} \right)_{avg}} = \frac{s}{t}\]

Substitute the value of $s$ and t in the above equation:

\[\begin{array}{c} {\left( {{v_{sp}}} \right)_{avg}} = \frac{{\left( {21\;{\rm{m}}} \right)}}{{\left( {10\;{\rm{s}}} \right)}}\\ = 2.1\;{\rm{m}}/{\rm{s}} \end{array}\]