Question

The particle starts from th origin with a velocity of 10ms-1and moves with a constant acceleration till the velocity increase to 50ms-1.at that instant ,the acceleration is suddenly reserved.what will be the velocity of the particle when it returns to the starting point

Answer 1

Answer:

70 m/s

Explanation:

DEAR FRIEND,

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THINKING PROCESS:

First, the particle is moving with a velocity of 10 m/s and then it increases from 10 m/s to 50 m/s. And then suddenly, the acceleration is reversed.

So, two cases can be considered, first of the change in velocity and then the sudden acceleration.

We can take out distances first, and then we can add them to find the velocity attained by the particle when it reaches the starting point knowing that the particle started from the rest i.e. the origin.

Formula which will be used is:  v² = u² - 2as.

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SOLVING APPROACH:

Let S be the distance travelled for the first time.

u = 10 m/s

v = 50 m/s

v² = u² - 2aS

(50)² = (10)² - 2aS

2500 = 100 - 2aS

S = $\frac{2400}{2a}$       .......(i)

Let s be the distance travelled for the second time.

u = 50 m/s

v = 0 m/s     ( ∵ the the particle has to reach to the starting point. )

v² = u² - 2as

0 = 2500 - 2as

s = $\frac{2500}{2a}$       .......(ii)

So,

Total Distance = S + s

                 s     = $\frac{2400}{2a}$ +  $\frac{2500}{2a}$

                 s     = $\frac{4900}{2a}$

Now,

At initial point, u = 0 m/s

v² = 2as

v² = 2a × $\frac{4900}{2a}$

v² = 4900

v  = $\sqrt{4900}$

v  = 70 m/s

CONCLUSION: The velocity of the particle when it returns at the starting point will be 70 m/s.

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CHEERS!!!

MARK AS BRAINLIEST!!!