Question

A body starts moving from rest and moves with a uniform acceleration of 5 cm/s?. In how much time the velocity of body will become 50 cm/s and how much will it travel in this line?

□Freë poïnts□​

Answer 1

Answer:

t=10s

d=2.5m

Step-by-step explanation:

in the question;the unit of acceleration is sec,that is incorrect.

the unit of acceleration is sec^2.

Answer 2

Given:

$\rightarrowtail \bold{Acceleration (a) = 5 cm/s}$

$\rightarrowtail \bold{Final\;velocity(v)=50cm/s}$

$\rightarrowtail \bold{Initial\;velocity(u)=0cm/s}$

∵ Car is moving from rest.

$\large\text{To find:}$

$\rightarrowtail\bold{Time\;taken(t)}$

$\rightarrowtail \bold{Distance\;covered(S)}$

$\large\text{Solution:}$

Using the first equation of motion.

$\mapsto\bold{v=u+at}$

Where we have.

$\twoheadrightarrow\bold{v=\dfrac{50}{100}m/s }$

$\twoheadrightarrow\bold{a=\dfrac{5}{100}m/s }$

$\twoheadrightarrow\bold{u=0m/s}$

Substituting we get.

$\nrightarrow \bold{0.5 = 0 + 0.05(t)}$

$\nrightarrow \bold{0.5=0.05t}$

$\nrightarrow\bold{t=\dfrac{0.5}{0.05} }$

∴ Time taken (t) is 10 seconds.

Using the second equation of motion.

$\mapsto \bold{S=ut+\dfrac{1}{2}at^{2} }$

Where we have.

$\twoheadrightarrow \bold{u=0m/s}$

$\twoheadrightarrow \bold{a=0.05m/s^{2}}$

$\twoheadrightarrow\bold{t=10sec}$

Substituting we get.

$\nrightarrow \bold{S=0(10)+\dfrac{1}{2}(0.05)(10)^{2} }$

$\nrightarrow \bold{S=0+0.025(100)}$

∴ The distance covered is 2.5 meters.

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