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

Answer:

0.0096 seconds

5208.33 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{50^2-0^2}{2\times 0.24}\\\Rightarrow a=5208.33\ m/s^2$

The acceleration of the firework shell was 5208.33 m/s²

$v=u+at\\\Rightarrow 50=0+5208.33t\\\Rightarrow t=\frac{50}{5208.33}=0.0096\ s$

The acceleration lasted for 0.0096 seconds

Answer 2

Answer:

(a) 0.0096 sec

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

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 $s=ut+\frac{1}{2}at^2$

$s=0\times t+\frac{1}{2}at^2=\frac{1}{2}at^2$

Using the value of at = 50

$0.240=\frac{1}{2}\times 50\times t$

$t=0.0096sec$

(b) As at = 50

$a\times 0.0096=50$

So $a=5208.333m/sec^2$