Physics, 14.02.2021 14:00 calthecrystal

A fireworks shell is accelerated from rest to a velocity of 56.0 m/s over a distance of 0.270 m. (a) How long (in s) did the acceleration last?

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Answer from: sean7900

(a) The acceleration lasts 9.642 seconds.

(b) The acceleration of the fireworks shell is 5.808 meters per square second.

Explanation:

Statement is incomplete. Complete description is presented below:

A fireworks shell is accelerated from rest to a velocity of 56 meters per second over a distance of 0.270 meters. (a) How long (in seconds) did the acceleration last? (b) Calculate the acceleration.

(b) Let suppose that the fireworks shell is accelerated uniformly. Given that initial ( $v_{o}$ ) and final speed ( $v_{f}$ ), measured in meters per second, and distance (), measured in meters, are known, we calculate the acceleration (), measured in meters per square second, by this kinematic expression:

$a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot s}$ (1)

If we know that $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 56\,\frac{m}{s}$ and $s = 0.270\,m$ , then the acceleration is:

$a = \frac{\left(56\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (270\,m)}$

$a = 5.808\,\frac{m}{s^{2}}$

The acceleration of the fireworks shell is 5.808 meters per square second.

(a) Now we calculate the time associated with acceleration (), measured in seconds, by means of this kinematic equation:

$t = \frac{v_{f}-v_{o}}{a}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 56\,\frac{m}{s}$ and $a = 5.808\,\frac{m}{s^{2}}$ , then time associated with acceleration is:

$t = \frac{56\,\frac{m}{s}-0\,\frac{m}{s}}{5.808\,\frac{m}{s^{2}} }$

$t = 9.642\,s$

The acceleration lasts 9.642 seconds.

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organizations that test products for safety are

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a national fire protection association

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A fireworks shell is accelerated from rest to a velocity of 56.0 m/s over a distance of 0.270 m. (a...

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