Question

A firework shell is accelerated from rest to a velocity of 50 m/s over a distance of 0.250 km. Calculate
the acceleration. How long did the acceleration last ?

CBSE Class IX Science SA 2 (3 Marks)

Answer 1

Answer:

t = 10 seconds

Explanation:

u = 0 m/s, v = 50 m/s, d = 0.250 km = 250 m

v² = u² + 2as

(50)² = (0)² + 2 x a x 250  

2500 = 0 + 500 x a  

2500 = 500 x a  

 a = 2500 ÷ 500  

a = 5 m/s²  

v = u + at

50 = 0 + 5 x t  

t = 50 ÷ 5  = 10

t = 10 seconds

Answer 2

Answer:

5$m/s^2$ is the required acceleration and acceleration last for 10 sec.

Explanation:

Given, a firework shell is accelerated from rest to a velocity of 50m/s over a distance 0.250km.

So, initial velocity (u) = 0 , final velocity (v) = 50m/s

and distance = 0.250km =  0.250 × 1000m = 250m.

Now, we know that the formula of motion , $v^2 - u^2 = 2as$.

On putting the given value in the above formula we get,

$(50)^2 - (0)^2 = 2a (250)$

⇒ a = $\frac{2500}{2 (250)}$ = 5m/$s^2$

So, the acceleration is 5$m/s^2$ .

Now, from the formula of  first equation of motion v = u + at.

⇒(50) = 0 + 5 (t)

⇒t = $\frac{50}{5}$ = 10 sec.

So, acceleration last for  10 sec.

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