Question

A particle moves in a straight line with a constant acceleration. it changes its velocity from 20m/S to 40m/s while passing through a distance 270m in t second.The value of t is​

Answer 1

Answer:

The time taken to pass through a distance of 135 m is 9 seconds.

Given

Initial velocity, u = 10 \mathrm{ms}^{-1}u=10ms

−1

Final velocity, v = 20 \mathrm{ms}^{-1}v=20ms

−1

Distance, s = 135 m

Solution:

Let a be the constant acceleration of the particle.

Equation

We know that the distance travelled by a body under constant acceleration is given by

s=\frac{v^{2}-u^{2}}{2 a}s=

2a

v

2

−u

2

\Rightarrow a=\frac{v^{2}-u^{2}}{2 s} \rightarrow(1)⇒a=

2s

v

2

−u

2

→(1)

We know that the final velocity of a body under constant acceleration is given by

v = u+at

Substituting equation (1) in the above equation, we get

v=u+\left[\frac{\left(v^{2}-u^{2}\right)}{2 s}\right] tv=u+[

2s

(v

2

−u

2

)

]t

Equating to time t, we get

t=\frac{[2 s(v-u)]}{v^{2}-u^{2}}t=

v

2

−u

2

[2s(v−u)]

t=\frac{2 s(v-u)}{(v-u)(v+u)}t=

(v−u)(v+u)

2s(v−u)

t=\frac{2 s}{v+u}t=

v+u

2s

Substitution

t=\frac{2 \times 135}{10+20}t=

10+20

2×135

t=\frac{270}{30}t=

30

270

t = 9 s"

Answer 2

Answer:

9 sec

Explanation:

we can use

${v}^{2} - {u}^{2} = 2as$

here we have v=40 u=20 s=270

to find a=?

$1600 - 400 = 2 \times a \times 270$

$a = \frac{20}{9}$

we can put this in

$v - u = at$

$40 - 20 = (20 \div 9) \times t$

$t = 9$