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Irina18 [472]
3 years ago
12

A body accelerate uniformly from rest at 0.2m/s for one-fifth of a minute. Calculate the distance covered by the body.​

Physics
1 answer:
PtichkaEL [24]3 years ago
4 0

Answer:

<h2>14.4 m</h2>

The distance covered by the body is <u>1</u><u>4</u><u>.</u><u>4</u><u> </u><u>m</u><u>.</u>

<u>Sol</u><u>ution</u><u>,</u>

Initial velocity(u)= 0 m/s( according to Question, the body starts from rest.that's why initial velocity(u) becomes zero.

Acceleration (a)= 0.2 m/s

Time (t):

\frac{1}{5} minute =  \frac{1}{5 }  \times 60 \: seconds

= 12 \: seconds

Now,

Applying third equations of motion:

s = ut +  \frac{1}{2} a {t}^{2}  \\ s = 0 \times 12 +  \frac{1}{2}  \times 0.2 \times  {(12)}^{2}  \\s = 0 + 14.4 \\ s = 14.4 \: metre

Thus, the distance covered by the body is 14.4 metre.

Further more information:

<h3><u>Application</u><u> </u><u>of</u><u> </u><u>equation</u><u> </u><u>of</u><u> </u><u>motion</u><u> </u><u>in</u><u> </u><u>differ</u><u>ent</u><u> </u><u>situa</u><u>tion</u><u>:</u></h3>
  • When a certain object comes in motion from rest, in the case, initial velocity(u)= 0 m/s
  • When a moving object comes in rest,in the case, final velocity(v)= 0 m/s
  • If the object is moving with uniform velocity, in the case, (u=v)
  • If any object is thrown vertically upward, in the case, acceleration (a)= -g
  • When an object is falling from certain height, in the case, u= 0 m/s
  • When an object is thrown vertically upwards in the case, final velocity at maximum height becomes 0.

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