Question: A fireworks shell is accelerated from rest to a velocity of\({\bf{65}}.{\bf{0}}{\rm{ m/s}}\)over a distance of\({\bf{0}}.{\bf{250}}\;{\bf{m}}\).

(a) How long did the acceleration last?

(b) Calculate the acceleration.

• Initial velocity U = \({\bf{0}}\;{\rm{m/s}}\).
• Final velocity V = \({\bf{65}}.{\bf{0}}\;{\rm{m/s}}\).
• Traveled distance D = \({\bf{0}}.{\bf{250}}\;{\bf{m}}\)

a)

The law of equation can calculate the time as:

\(V = U + at\)

Here V is the final velocity, U is the initial velocity, a is the acceleration, and t ia the time.

Substituting values in the above expression, we get,

\(\begin{array}{c}65 = 0 + (8450) \times t\\t = 0.0076\;s\end{array}\)

The time taken by the firework shell is \({\bf{0}}.{\bf{0076}}\;{\rm{s}}\).

b)

From the equation of motion, we can calculate the acceleration of the system if we have initial and final velocity and the distance traveled.

The equation can be written as:

\({V^2} - {U^2} = 2ad\)

Here V is the final velocity, U is the initial velocity, a is the acceleration, and d is the traveled distance.

Substituting values in the above expression, we get,

\(\begin{array}{c}{(65)^2} - {(0)^2} = 2 \times a \times (0.25)\\4225 = 2 \times a \times (0.25)\\a = \frac{{4225}}{{2 \times 0.25}}\\a = 8450\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\end{array}\)

Hence the acceleration obtained is \(8450\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\).