Question

QUESTION 1:

A fireworks shell is accelerated from rest to a velocity of 55.0 m/s over a distance of 0.210 m.

(a)

How long (in s) did the acceleration last?

s

(b)

Calculate the acceleration (in m/s²). (Enter the magnitude.)

m/s²

^{QUESTION 2:}

A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 5.70 ✕ 10⁵ m/s² for 9.60 ✕ 10⁻⁴ s. What is its muzzle velocity (in m/s) (that is, its final velocity)? (Enter the magnitude.)

Answer 1

Hi,

Hope you are doing well.

QUESTION 1:

I'm doing Part (b) first and then Part (a) which is more convenient.

PART (b):

We know that, according to kinematic equations, final speed is given by,

Where we have,

Final speed,

Initial speed,

Distance traveled,

Hence acceleration of the shell is given by,

PART (a):

We know that, according to kinematic equations, final speed is given by,

Where we have,

Final speed,

Initial speed,

Acceleration of the shell,

Hence time taken by the shell is given by,

QUESTION 2:

We know that, according to kinematic equations, final speed is given by,

Where we have,

Acceleration,

Initial speed,

Time taken,

Hence the final speed is given by,

​​​​​​​

Answer 2

QUESTION 1: (a) Using the kinematic equation: v^2 = u^2 + 2as where u=0 (initial velocity), v=55.0 m/s (final velocity), s=0.210 m (distance), we can solve for the time duration of acceleration: t = (v^2 - u^2) / 2as = (55.0^2) / (2*0.210) = 732.14 ms = 0.73214 s (to 5 significant figures)

Therefore, the acceleration lasted for 0.73214 seconds.

(b) The acceleration (a) can be calculated using the same kinematic equation: a = (v^2 - u^2) / 2s = (55.0^2) / (2*0.210) = 7230.95 m/s^2 ≈ 7231 m/s^2 (to 4 significant figures)

Therefore, the acceleration of the fireworks shell was approximately 7231 m/s^2.

QUESTION 2: Using the kinematic equation: v = u + at where u=0 (initial velocity), a=5.70 ✕ 10^5 m/s^2 (acceleration), and t=9.60 ✕ 10^-4 s (time), we can solve for the final velocity (v): v = u + at = 0 + (5.70 ✕ 10^5)(9.60 ✕ 10^-4) = 547.2 m/s (to 3 significant figures)

Therefore, the muzzle velocity of the bullet is 547.2 m/s.