Question

the velocity of car becomes twice the previous velocity after every 2 seconds. If the average velocity of car after 6 second is 14m/s determine the distance travelled by car in first 4 seconds.

Answer with explaination​

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

Thank you!

Hope it helps!

Answer 2

ans

35

explaination

At 6 seconds  velocity= 14ms^-1

At 4 seconds it will equal to = 7ms^-1

At 2 it will equal to = 3.5ms^-1

At u it will equal to = 1.75ms^-1

acceleration = (v-u)/t

(14-7)/2

a=3.5

so by using

s=ut+0.5at^2

s=1.75*4+0.5*3.5*4*4

s=35