Question

A particle moving with uniform acceleration travels 24 metres and 64 metres in first two successive intervals of 4 seconds each. Its initial velocity is ?

Answer 1

Answer:

Explanation:

Given :-

Distance traveled in 1st 4 seconds = 24 m

And next 4 seconds = 64 m

Time taken in two intervals, t = 4 seconds each

To Find :-

Initial velocity, u = ??

Formula to be used :-

1st and 2nd equation of motion,

v = u + at and s = ut + 1/2 at²

Solution :-

Let initial velocity of particle be u m/s.

We know that,

v = u + at

⇒ v = u + a × 4

⇒ v = u + 4a

Now, s = ut + 1/2 at²

⇒ 24 = 4u + 1/2 × a × 4 × 4

⇒ 24 = 4u + 8a

⇒ u + 2a = 6 .... (i)

Now, for next 4 seconds,

s = ut + 1/2 at²

⇒ 6u = (u + 4a) × 4 + 1/2 × a × 16  [From (i)]

⇒ 16 = u + 4a + 2a

⇒ u + 6a = 16 ..... (ii)

Now, From Eq (i) and (ii), we get

 u + 6a = 16

 u + 2a = 6

-    -         -

__________

⇒ 4a = 10

⇒ a = 10/4

⇒ a = 2.5

Now putting a's value in Eq (i), we get

⇒ u + 2a = 6

⇒ u + 2 × 2.5 = 6

⇒ u + 5 = 6

⇒ u = 6 - 5

⇒ u = 1 m/s

Hence, the initial velocity of particle is 1 m/s.

Answer 2

$\red{ \sf{Answer}}$

1 m per second

$\red{ \sf{Formula\:\: used }}$

• $s = ut + \frac{1}{2} a {t}^{2}$

$</p><p> \red{ \sf{Figure }}$

• see attachment

$</p><p> \red{ \sf{Solution}}$

$\rightarrow$Let initial velocity be u m per sec and acceleration be a m per square sec

since it is uniform motion so acceleration is constantly throughout.

$\rightarrow$

we divide motion in two parts

first motion is of 24 km

and then second is of 88 km ( total motion)

for first

$\rightarrow$$24 = 4u+ \frac{1}{2} a {4}^{2}$

$\rightarrow$6 = u +2a

$\rightarrow$12 = 2u +4a ............1

$\rightarrow$second motion is of 88 km ( 24 + 64 )

for it total time is 8 seconds

$\rightarrow$$88 =8u+ \frac{1}{2} a {8}^{2}$

$\rightarrow$11 = u + 4a ........... 2

sub 2 from 1

$\rightarrow$1 = u

so initial velocity is 1 m per sec