Question

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.

Short answer

a) \({\bf{0}}.{\bf{0076}}\;{\rm{s}}\).

b) \(8450\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\).

Step-by-step solution

Given data

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

Determining time

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 \({\bf{0}}.{\bf{0076}}\;{\rm{s}}\).

Determining Acceleration

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 \(8450\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\).