Question

A fireworks shell is accelerated from rest to a velocity of 50.0 m/s over a distance of 0.240 m. (a) How long (in s) did the acceleration last? ________ s (b) Calculate the acceleration in m/s^2). (Enter the magnitude.) _________m/s^2

Answer 1

(a) 0.0096 sec

(b)
`a=5208.333m/sec^2`

Step-by-step explanation:

We have given that firewall shell starts from rest so initial velocity u = 0 m/sec

And final velocity v = 50 m/sec

Distance s = 0.240 m

(A) From first equation pf motion we know that v = u+at

So 50 = 0+at

at = 50

Now from second equation of motion
`s=ut+(1)/(2)at^2`

`s=0* t+(1)/(2)at^2=(1)/(2)at^2`

Using the value of at = 50

`0.240=(1)/(2)* 50* t`

`t=0.0096sec`

(b) As at = 50

`a* 0.0096=50`

So
`a=5208.333m/sec^2`

Answer 2

0.0096 seconds

5208.33 m/s²

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

`v^2-u^2=2as\\\Rightarrow a=(v^2-u^2)/(2s)\\\Rightarrow a=(50^2-0^2)/(2* 0.24)\\\Rightarrow a=5208.33\ m/s^2`

The acceleration of the firework shell was 5208.33 m/s²

`v=u+at\\\Rightarrow 50=0+5208.33t\\\Rightarrow t=(50)/(5208.33)=0.0096\ s`

The acceleration lasted for 0.0096 seconds