Question

Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed and time are classified). What is its average acceleration in m/s2 and in multiples of g (\({\bf{9}}{\bf{.80}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\))?

Short answer

The acceleration of missile is \({\bf{108}}{\bf{.33}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\) or \({\bf{11}}{\bf{.05}}\,{\bf{g}}\).

Step-by-step solution

Determination of formula for average acceleration of missile

Given data:

The suborbital speed of missile is\(v = 6.50\;{\rm{km}}/{\rm{s}}\).

The initial speed of rocket is\(u = 0\).

The time for acceleration of the missile is\(t = 60\;{\rm{s}}\).

The acceleration of the missile is the increase in speed over time while moving from one position to another position.

The average acceleration of the missile is given by

\(a = \frac{{v - u}}{t}\)

Here, \(a\) is the acceleration of missile.

Determination of average acceleration of missile

Substitute all the values in the above equation.

\(\begin{array}{l}a = \frac{{\left( {6.50\;{\rm{km}}/{\rm{s}} - 0} \right)\left( {\frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)}}{{60\;{\rm{s}}}}\\a = 108.33\;{\rm{m}}/{{\rm{s}}^2}\\a = \left( {108.33\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {\frac{g}{{9.80\;{\rm{m}}/{{\rm{s}}^2}}}} \right)\\a = 11.05g\end{array}\)

Therefore, the acceleration of the missile is \(108.33\;{\rm{m}}/{{\rm{s}}^2}\) or \(11.05\,{\rm{g}}\).