Step 1

We are given the velocity ${v_0} = 300\;{\rm{ft/s}}$.

We are asked to determine the time when the jet drops and construct the v-t and s-t graph.

 
Step 2

To find the acceleration from graph we will use the relation,

\[\begin{array}{c} \frac{{a - {t_2}}}{{t - {a_1}}} = \frac{{{t_1} - {t_2}}}{{{a_2} - {a_1}}}\\ \frac{{a - \left( { - 20} \right)}}{{t - 10}} = \frac{{ - 10 - \left( { - 20} \right)}}{{20 - 10}}\\ a = \left( {t - 30} \right)\;{\rm{ft/}}{{\rm{s}}^2} \end{array}\]

To find the velocity we will use the relation,

\[\begin{array}{c} \int_{{v_0}}^v {dv} = \int_{10\,\;{\rm{s}}}^t {adt} \\ \int_{300}^v {dv} = \int_{10\,\;{\rm{s}}}^t {\left( {t - 30} \right)dt} \\ \left( {v - 300} \right) = \left[ {\frac{1}{2}{t^2} - 30t} \right]_{10\;{\rm{s}}}^t\\ \left( {v - 300} \right) = \left[ {\left( {\frac{1}{2}{t^2} - 30t} \right) - \left( {\frac{1}{2}{{\left( {10} \right)}^2} - 30\left( {10} \right)} \right)} \right]\\ v = \frac{1}{2}{t^2} - 30t + 550\;{\rm{ft/s}} \end{array}\]…… (1)

On plugging the value $t = 20\,{\rm{s}}$ in the above relation, we get,

\[\begin{array}{l} v = \left[ {\frac{1}{2}{{\left( {20} \right)}^2} - 30\left( {20} \right) + 550} \right]\;{\rm{ft/s}}\\ v = 150\;{\rm{ft/s}} \end{array}\]
 
Step 3

On using the initial conditions we will calculate the time at speed $v = 0$.

\[\begin{array}{c} \int_{150}^v {dv} = \int_{20\,\;{\rm{s}}}^t {a'dt} \\ \int_{150}^v {dv} = \int_{20\,\;{\rm{s}}}^t { - 10dt} \\ \left( {v - 150} \right) = \left[ { - 10t} \right]_{20\;{\rm{s}}}^{t'}\\ \left( {v - 150} \right) = \left[ { - 10t' - \left( { - 10 \times 20} \right)} \right]\\ v = \left( { - 10t' + 350} \right)\;{\rm{ft/s}} \end{array}\]

On plugging the values in the above relation, we get,

\[\begin{array}{c} 0 = \left( { - 10t' + 350} \right)\\ t' = 35\;{\rm{s}} \end{array}\]

To find the distance we will use the relation,

\[\begin{array}{c} \int_0^s {ds} = \int_0^t {{v_0}dt} \\ \int_0^s {ds} = \int_0^t {\left( {300} \right)dt} \\ s = 300t \end{array}\]

On plugging the value $t = 10\,{\rm{s}}$ in the above relation, we get,

\[\begin{array}{l} s = \left( {300 \times 10} \right)\;{\rm{ft}}\\ s = 3000\;{\rm{ft}} \end{array}\]
 
Step 4

On using the initial conditions $s = 3000\;{\rm{ft}}$ and $t = 10\,{\rm{s}}$ and equation (1) we will calculate the distance.

\[\begin{array}{c} \int_{3000}^s {ds} = \int_{10\;{\rm{s}}}^t {vdt} \\ \int_{3000}^s {ds} = \int_{10\;{\rm{s}}}^t {\left[ {\frac{1}{2}{t^2} - 30t + 550} \right]dt} \\ \left( {s - 3000} \right) = \left[ {\frac{1}{6}{t^3} - 15{t^2} + 550t} \right]_{10\;{\rm{s}}}^t\\ \left( {s - 3000} \right) = \left[ {\left( {\frac{1}{6}{t^3} - 15{t^2} + 550t} \right) - \left( {\frac{1}{6}{{\left( {10} \right)}^3} - 15{{\left( {10} \right)}^2} + 550\left( {10} \right)} \right)} \right]\\ s = \left( {\frac{1}{6}{t^3} - 15{t^2} + 550t - 1167} \right)\;{\rm{ft}} \end{array}\]

On plugging the value $t = 20\,{\rm{s}}$ in the above relation, we get,

\[\begin{array}{l} s = \left( {\frac{1}{6}{{\left( {20} \right)}^3} - 15{{\left( {20} \right)}^2} + 550\left( {20} \right) - 1167} \right)\;{\rm{ft}}\\ s = 5167\;{\rm{ft}} \end{array}\]

To find the distance at $t = 35\;{\rm{s}}$ we will use the relation,

\[\begin{array}{c} \int_{5167}^s {ds} = \int_{20\;{\rm{s}}}^t {\left( v \right)dt} \\ \int_{5167}^s {ds} = \int_{20\;{\rm{s}}}^t {\left[ {\left( { - 10t' + 350} \right)} \right]dt} \\ \left( {s - 5167} \right) = \left[ { - 5t{'^2} + 350t'} \right]_{20\;{\rm{s}}}^t\\ \left( {s - 5167} \right) = \left[ {\left( { - 5{t^2} + 350t} \right) - \left( { - 5{{\left( {20} \right)}^2} + 350 \times 20} \right)} \right]\\ s = \left( { - 5{t^2} + 350t + 167} \right)\;{\rm{ft}} \end{array}\]

On plugging the value $t = 35\,{\rm{s}}$ in the above relation, we get,

\[\begin{array}{l} s = \left( { - 5{{\left( {35} \right)}^2} + 350\left( {35} \right) + 167} \right)\;{\rm{ft}}\\ s = 6292\;{\rm{ft}} \end{array}\]
 
Step 5

The following is the graph between speed and time.

The following is the graph between distance and time.