Question

A particle having initial velocity 10 m/s moves witha

constant acceleration 5ms-², for a time 15 second along
a straight line, what is the displacement of the particle in
the last 2 second?


​

Answer 1

Explanation:

displacement in 15 second

s=ut+1/2at^2

s=10×15+5/2×225

=1425/2

displacement in 13 second

s=10×13+5/2×169

s=130+5/2×169

s=1105/2

displacement in 15 second-displment I 13 second

1425/2-1105/2=320/2=160

Answer 2

The displacement of the particle in the last two seconds is equal to 160 m.

Explanation:

According to the given information, it is given that, a particle having initial velocity 10 m/s moves with a constant acceleration 5m$s^{-2}$, for a time 15 second along a straight line.

This means that, the initial velocity of the particle that is u is 10, the acceleration of the particle along the straight line that is a is 5 and the time taken by the particle that is t is 15 seconds.

Now, we know that, the well - known equation of motion is,

S is equal to ( ut + $\frac{1}{2} at^{2}$ ) , where S is the displacement of the particle.

Now, for a time period of 15 seconds, the displacement $S_{1}$ is equal to

$S_{1} = (10 * 15) + \frac{1}{2} *(5*15^{2})$

Or, $S_{1} = 150 + 562.5 = 712.5$ m.

Now, for a time period of 13 seconds, the displacement $S_{2}$ is equal to

$S_{2} = (10 * 13) + \frac{1}{2} *(5*13^{2})$

or, $S_{2} = 130 + 422.5 = 552.5$ m.

Now, the displacement of the particle in the last two seconds is, the difference of the values of $S_{2}$ and $S_{1}$.

Then, the displacement of the particle in the last two seconds is equal to,

712.5 - 552.5 = 160 m.

Thus, the displacement of the particle in the last two seconds is equal to 160 m.

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