Question

An object starts moving from rest with uniform acceleration and travels 20m in first 2seconds . what will be its velocity after 10s from the start and displacement of the particle in 10th second

Answer 1

NOTE - am not sure of which laws comes in which number, but am sure of the laws

Answers:

1. 100m/s
2. 135m

Explanation:

The object starts to move from rest so,

u (initial velocity) = 0

s (displacement) = 20m

t (time) = 2secs

so using these values we can use the 2nd law of motion;

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

20 m = 1/2 (a)(2)^2

20*2/4 = a

10m/s^2 = a

so we got the acceleration.

1. to find the the velocity after 10secs

u = 0 (as it has been mentioned "from start")

t = 10sec

a (which we have found out) = 10m/s^2

so we use the 1st law;

$v = u + at$

and by plugging in the values,

v = 100m/s

  2. to find the displacement of the particle at 10th second

we use the special formula:-

$S_{n^{th} } = u + \frac{1}{2}a(2n - 1)$

For u:

$S = ut + \frac{1}{2}at^{2}$   (all this we do for till 9sec because final velocity at (n-1)sec =     initial velocity at nsec)

u = 0

a = 10

t = 9sec

we get s = 405m

then we use another formula:

$v^{2} = u^{2} + 2aS$

we get v = 90 = u(for 10th sec)

so then we use the special one:

$S_{n^{th} } = u + \frac{1}{2}a(2n - 1)$

to get $S_{n^{th} }$ = 135 m

oof!!! sum was so big!!!

now rejoice!!!!