Question

In 3.0 s, a particle moving with constant acceleration along the x axis goes from x = 10 m to x = 30 m. The velocity at the end of this time interval is 15 m/s. What is the acceleration of the particle?

Answer 1

Given info : In 3.0 s, a particle moving with constant acceleration along the x axis goes from x = 10 m to x = 30 m. The velocity at the end of this time interval is 15 m/s.

To find : the acceleration of the particle.

solution : let initial velocity is u

Distance covered by particle is s = 30 - 10 = 20m

final velocity, v = 15 m/s

Time taken, t = 3 sec

using formula, v = u + at

⇒15 = u + 3a .......(1)

Now using, s = ut + 1/2 at²

⇒20 = 3u + 1/2 × a × 3²

⇒20 = 3u + 4.5a ......(2)

From equations (1) and (2)

3 × 15 - 20 = 3(u + 3a) - 3u - 4.5a

⇒25 = 9a - 4.5a

⇒a = 25/4.5 = 5/0.9 =50/9 = 5.55 m/s²

Therefore the acceleration of particle is 5.55 m/s²

Answer 2

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It is given that :-

• Distance travelled by particle = 30 -10 = 20m
• Initial velocity = u = ?
• Final velocity = 15m/s
• Time = 3s
• Acceleration = ?

Now, we know according to first equation of motion :

$v = u + at$

After inputting the known values we get :-

$15 = u + 3a$

We also know according to second equation of motion :

$s = ut + \frac{1}{2} \times a {t}^{2}$

After inputting the known values we get :-

$2 0 = 3u + \frac{1}{2} \times {3}^{2} a$

$20 = 3u + \frac{9a}{2} = 20 = 3u + 4.5a$

From these both equations we can say:

$3 \times 15 - 20 = 3(u + 3a) - 3u - 4.5a$

$25 = 9a - 4.5a$

$25 = 4.5a$

$a = \frac{25}{4.5} = \cancel \frac{25}{4.5}$

$a = 5.55$

Therefore we get the acceleration of particle to be 5.55m/s^2.

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BY SCIVIBHANSHU

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