Question

A particle moves in a straight line with a constant acceleration. It changes its velocity from10m/s to 20m/s while passing through distance 135m in t seconds, the value of t is

Answer 1

U=10m/s
V=20m/s
S=135m
We know,
2as=v2-u2
2×a×135=(20×20)-(10×1₩)
270a=400-100
a=300/270
a=10/9m/s2
V=u+at
20=10+10/9×t
10=10/9×t
T=10×9/10
T=9 seconds
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Answer 2

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

Given

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

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

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

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

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

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

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

Equating to time t, we get

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

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

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

Substitution

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

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

t = 9 s"