Question

If a rock is thrown upward on the planet Mars with a velocity of\({\bf{10 m/s}}\), its height (in meters) after\(t\)seconds is given by\(H = 10t - 1.86{t^2}\).

(a) Find the velocity of the rock after one second.

(b) Find the velocity of the rock when\(t = a\).

(c) When will the rock hit the surface?

(d) With what velocity will the rock hit the surface?

Short answer

(a)The velocity of the rock after one secondis \(6.28\;{\rm{m/sec}}\).

(b)The velocity of the rock when \(t = a\)is\(\left( {10 - 3.72a} \right)\;{\rm{m/sec}}\).

(c)The rock will hit the surface when \(t = 5.4\;{\rm{s}}\).

(d)The velocity at which the rock hit the surface is\( - 10{\rm{ m/s}}\).

Step-by-step solution

(a) Step 1: Find the velocity of the rock after one second

Apply the definition of instantaneous velocity which states that the position function \(f\left( t \right)\) at time \(t = a\) is represented as \(v\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\).

Now find the velocity at time \(t = 1\;{\rm{seconds}}\).

\(\begin{aligned}v\left( 1 \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{H\left( {1 + h} \right) - H\left( 1 \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\left( {10\left( {1 + h} \right) - 1.86{{\left( {1 + h} \right)}^2}} \right) - \left( {10 - 1.86} \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{10 + 10h - 1.86\left( {1 + 2h + {h^2}} \right) - 10 + 1.86}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{10 + 10h - 1.86\left( {1 + 2h + {h^2}} \right) - 10 + 1.86}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{10 + 10h - 1.86 - 3.72h - 1.86{h^2} - 10 + 1.86}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \left( {6.28h - 1.86h} \right)\\ &= 6.28\end{aligned}\)

Thus, the velocity of the rock after one second is \(6.28\;{\rm{m/sec}}\).

(b) Step 2: Find the velocity of the rock at \(t = a\)

Apply the definition of instantaneous velocity which states that the position function \(f\left( t \right)\) at time \(t = a\) is represented as \(v\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\).

Now find thevelocityat time\(t = a\).

\(\begin{aligned}v\left( a \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{H\left( {a + h} \right) - H\left( a \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\left( {10\left( {a + h} \right) - 1.86{{\left( {a + h} \right)}^2}} \right) - \left( {10a - 1.86{a^2}} \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{10a + 10h - 1.86\left( {{a^2} + 2ah + {h^2}} \right) - 10a + 1.86{a^2}}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{10a + 10h - 1.86{a^2} - 3.72ah - 1.86{h^2} - 10a + 1.86{a^2}}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{10h - 3.72ah - 1.86{h^2}}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \left( {10h - 3.72a - 1.86h} \right)\\ &= 10 - 3.72a\end{aligned}\)

Thus, the velocity of the rock when \(t = a\) is \(\left( {10 - 3.72a} \right)\;{\rm{m/sec}}\).

(c) Step 3: Find when the rock will hit the surface

Put\(H = 0\)into\(H = 10t - 1.86{t^2}\).

\(\begin{aligned}0 &= 10t - 1.86{t^2}\\t\left( {10 - 1.86t} \right) &= 0\\t &= 0\end{aligned}\)

Or,

\(\begin{aligned}10 - 1.86t &= 0\\t &= \frac{{10}}{{1.86}}\\ &= 5.4\end{aligned}\)

Thus, the rock will hit the surface when \(t = 5.4\;{\rm{s}}\).

(d) Step 4: Find the velocity at which the rock hit the surface

Consider the velocity\(\left( {10 - 3.72a} \right)\;{\rm{m/sec}}\).

\(\begin{aligned}v\left( {5.4} \right) &= 10 - 3.72\left( {5.4} \right)\\ &= - 10{\rm{ m/s}}\end{aligned}\)

Thus, the velocity at which the rock hit the surface is\( - 10{\rm{ m/s}}\).