Step 1

We are given the following data:

The equation of $T$ is $T = \left( {4{t^{\frac{1}{2}}}} \right)\;{\rm{kN}}$.

The value of time is $t = 35\;{\rm{s}}$.

The initial velocity of the rocket is ${v_1} = 0$. (Since start from rest)

We are asked to determine the sleds maximum velocity and the distance the sled travels when $t = 35\;{\rm{s}}$.

 
Step 2

We will draw the free-body diagram of the rocket sled.

Here, $N$ is the normal reaction force and $v$ is the velocity of the rocket.

The unit of mass of the rocket can be converted into SI units as,

\[\begin{array}{c} m = 4\;{\rm{Mg}}\\ m = \left( {4\;{\rm{Mg}}} \right)\left( {\frac{{{{10}^6}{\rm{g}}}}{{1\;{\rm{Mg}}}} \times \frac{{{{10}^{ - 3}}\;{\rm{kg}}}}{{1\;{\rm{g}}}}} \right)\\ m = 4000\;{\rm{kg}} \end{array}\]
 
Step 3

For, $0 \le t < \;25\;{\rm{s}}$,

We will apply the impulse momentum principle,

\[\begin{array}{c} \left[ {m{v_1} + \sum {\int_{{t_1}}^{{t_2}} {Fdt} } } \right] = mv\\ \left[ {m{v_1} + \left( {\int_{{t_1}}^{{t_2}} {Tdt} } \right)} \right] = mv \end{array}\]

Here, the limit of integration can be considered as, ${t_1} = 0\;{\rm{s}}$ to ${t_2} = 25\;{\rm{s}}$.

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ \begin{array}{l} \left( {4000\;{\rm{kg}}} \right)\left( {0\;{\rm{m/s}}} \right)\\ + \int_{0\;{\rm{s}}}^{25\;{\rm{s}}} {\left\{ {\left( {4{t^{\frac{1}{2}}}\;{\rm{kN}}} \right)\;\left( {\frac{{{{10}^3}\;{\rm{N}}}}{{1\;{\rm{kN}}}}} \right)} \right\}dt} \end{array} \right] = \left( {4000\;{\rm{kg}}} \right)\left( v \right)\\ \left( {4000\,{\rm{N}}} \right)\left[ {\frac{2}{3}{{\left( t \right)}^{\frac{3}{2}}}} \right]_{0\;{\rm{s}}}^{25\;{\rm{s}}} = \left( {4000\;{\rm{kg}}} \right)\left( v \right)\\ \left( {4000\;{\rm{N}}} \right)\left[ {\frac{2}{3}{{\left( {25} \right)}^{\frac{3}{2}}} - 0} \right] = \left( {4000\;{\rm{kg}}} \right)\left( v \right)\\ v = 83.33\;{\rm{m/s}} \end{array}\]
 
Step 4

From the given graph between $T\left( {{\rm{kN}}} \right)$ and $t$,

For $25\;{\rm{s}}\;{\rm{ < }}\;{\rm{t}}\;{\rm{ < }}\;{\rm{35}}\;{\rm{s}}\;$, the equation of $T$ will be formed as,

\[\begin{array}{c} \frac{{\left( {T - 0} \right)\;{\rm{kN}}}}{{\left( {t - 35\;{\rm{s}}} \right)}} = \frac{{\left( {20\;{\rm{kN}}\; - 0\;{\rm{kN}}} \right)}}{{\left( {25\;{\rm{s}}\; - 35\;{\rm{s}}} \right)}}\\ \frac{{\left( {T - 0} \right)\;{\rm{kN}}}}{{\left( {t - 35\;{\rm{s}}} \right)}} = \frac{{20\;{\rm{kN}}}}{{\left( { - 10\;{\rm{s}}} \right)}}\\ T = \left( {2\;{\rm{kN}}\; \times \frac{{{{10}^3}\;{\rm{N}}}}{{1\;{\rm{kN}}}}} \right)\left( {35\;{\rm{s}}\; - t} \right)\\ T = \left( {2 \times {{10}^3}{\rm{N}}} \right)\left( {35\;{\rm{s}}\; - t} \right) \end{array}\]
 
Step 5

We will apply the impulse momentum principle,

\[\begin{array}{c} \left[ {mv + \sum {\int_{{t_1}}^{{t_2}} {Fdt} } } \right] = m{v_2}\\ \left[ {mv + \left( {\int_{{t_1}}^{{t_2}} {Tdt} } \right)} \right] = m{v_2} \end{array}\]

Here, the limit of integration can be considered as, ${t_1} = 25\;{\rm{s}}$ to ${t_2} = t$ and ${v_2}$ represent the maximum velocity of rocket.

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ \begin{array}{l} \left( {4000\;{\rm{kg}}} \right)\left( {83.33\;{\rm{m/s}}} \right)\\ + \int_{25\;{\rm{s}}}^t {\left\{ {\left( {2 \times {{10}^3}\;{\rm{N}}} \right)\left( {35\;{\rm{s}}\; - t} \right)} \right\}dt} \end{array} \right] = \left( {4000\;{\rm{kg}}} \right)\left( {{v_2}} \right)\\ \left[ {\left( {3.33 \times {{10}^5}\;{\rm{N}}} \right) + \left( {2 \times {{10}^3}\,{\rm{N}}} \right)\left[ {35t - \frac{{{t^2}}}{2}} \right]_{25\;{\rm{s}}}^t} \right] = \left( {4000\;{\rm{kg}}} \right)\left( {{v_2}} \right) \end{array}\]

Substitute the limits in t.

\[\begin{array}{c} \left\{ \begin{array}{l} \left( {3.33 \times {{10}^5}\;{\rm{N}}} \right)\\ + \left( {2 \times {{10}^3}\,{\rm{N}}} \right)\left[ \begin{array}{l} 35t - \frac{{{t^2}}}{2} - \\ 35\left( {25} \right) + \frac{{{{\left( {25} \right)}^2}}}{2} \end{array} \right] \end{array} \right\} = \left( {4000\;{\rm{kg}}} \right)\left( {{v_2}} \right)\\ {v_2} = \left( { - 0.25{t^2} + 17.5t - 197.9167} \right)\;{\rm{m/s}} \end{array}\]
 
Step 6

Substitute $t = 35\;{\mathop{\rm s}\nolimits} $ in the equation ${v_2} = \left( { - 0.25{t^2} + 17.5t - 197.9167} \right)\;{\rm{m/s}}$ to obtain maximum velocity.

\[\begin{array}{c} {\left( {{v_2}} \right)_{{\rm{max}}}} = \left[ { - 0.25{{\left( {35} \right)}^2} + 17.5\left( {35} \right) - 197.9167} \right]\;{\rm{m/s}}\\ {\left( {{v_2}} \right)_{{\rm{max}}}} = 108.33\;{\rm{m/s}} \end{array}\]
 
Step 7

According to the kinematics principle, the displacement of the sled can be obtained by integrating the equation $ds = vdt$. For $0 \le t < \left( {25\;{\rm{s}}} \right)$, the condition is $s = 0\;{\rm{m}}$ at $t = 0\;\sec $.

The equation to calculate the displacement of the sled is given by,

\[\int_0^s {ds} = \int_{{t_1}}^{{t_2}} {vdt} \]

Here, $s$ represent the displacement of the sled and the limit of integration will be considered as ${t_1} = 0\;{\rm{s}}$ to ${t_2} = 25\;{\rm{s}}$.

Substitute all the known values in the above equation.

\[\begin{array}{c} \int_0^s {ds} = \int_0^{25\;{\rm{s}}} {\left( {\frac{2}{3}{{\left( t \right)}^{\frac{3}{2}}}} \right)dt} \\ \left[ s \right]_0^s = \left( {\frac{2}{3}} \right)\left[ {\frac{{2{{\left( t \right)}^{\frac{5}{2}}}}}{5}} \right]_0^{25\;{\mathop{\rm s}\nolimits} }\\ \left[ {s - 0} \right] = \left( {\frac{4}{{15}}} \right)\left[ {{{\left( {25} \right)}^{\frac{5}{2}}} - 0} \right]\\ s = 833.33\;{\rm{m}} \end{array}\]
 
Step 8

For, $\left( {25\;{\rm{s}}} \right)\; < t \le \left( {35\;{\rm{s}}} \right)$, the initial condition is $s = 833.33\;{\rm{m}}$ at $t = 25\;{\rm{s}}$.

The equation to calculate the displacement of the sled is given by,

\[\int_{{s_1}}^{{s_2}} {ds} = \int_{{t_1}}^{{t_2}} {vdt} \]

Here, $s$ represent the displacement of the sled and the limit of integration will be considered as ${t_1} = 0\;{\rm{s}}$ to ${t_2} = 25\;{\rm{s}}$.

Substitute all the known values in the above equation.

\[\begin{array}{c} \int_{833.33\;{\rm{m}}}^s {ds} = \int_{25\;{\rm{s}}}^{35\;{\rm{s}}} {\left( \begin{array}{l} - 0.25{t^2} + 17.5t\\ - 197.9167 \end{array} \right)dt} \\ \left[ s \right]_{\left( {833.33\;{\rm{m}}} \right)}^s = \left[ \begin{array}{l} - 0.08333{t^3} + 8.75{t^2}\\ - 197.9167t \end{array} \right]_{25\;{\rm{s}}}^{35\;{\mathop{\rm s}\nolimits} }\\ \left[ {s - 833.33\;{\rm{m}}} \right] = \left[ \begin{array}{l} - 0.08333{\left( {35} \right)^3} + 8.75{\left( {35} \right)^2} - 197.9167\left( {35} \right) + \\ 0.08333\left( {25} \right) - 8.75{\left( {25} \right)^2} + 197.9167\left( {35} \right) \end{array} \right]\\ s = 1833.33\;{\rm{m}} \end{array}\]