Question

.The velocity of car becomes twice the previous velocity after every two seconds. If the average velocity
of car after 6 seconds is 14 m/s, determine the distance travelled by car in first 4 seconds.
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Answer 1

Answer:

62.5 m

Explanation:

Let car's initial velocity be v

then after 2 seconds velocity=2v

so, using equation of motion

2v=v+2a (I equation of motion)

a=v/2

now average velocity after 6 seconds=14 m/s

therefore, 14=displacement/6

displacement in 6 seconds=84m (1)

now distance travelled in first four seconds=v×4 + 1/2×v/2×(4)^2 (2)

(II equation of motion)

using equation (1)

84=v×6 + 1/2×v/2×(6)^2

On solving,we get v=75/6m/s

putting the value of v in equation (2)

we get, s=375/6=125/2=62.5 m

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