Question

D. If the velocity of particle change from 4m/sec to 10 meter/sec in 3 sec then calculate th
distance of a particle covered in 10 seconds.​

Answer 1

Answer:

If the velocity of the particle changes at a constant rate, then this rate is called the constant acceleration. ... If the particle has a velocity of 4 m/s initially (at t=0) and has a constant ... rate of change) from 5 m/s to 2 m/s over one second, its constant acceleration is −3 m/s2. ... After one second, the car's velocity is 2509−10 m/s.

Explanation:

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Answer 2

$\huge \mathbb{\underbrace{\red{Question\:}}}$

If the velocity of particle change from 4m/s to 10 m/s in 3 sec then calculate the distance of a particle covered in 10 seconds.

$\huge \mathbb{\underbrace{\red{Given\:}}}$

u = 4m/s ; v = 10m/s ; t = 3 ; s = ?

$\huge \mathbb{\underbrace{\red{Answer\:}}}$

v = u + at

10 = 4 + 3a

10 - 4 = 3a

3a = 6

a = 2m/s²

Now,

We have to find the value of distance(s)

s = ut + 1/2 at²

s = 4 × 3 + 1/2 × 2 × 3 × 3

s = 12 + 9

s = 21m

In 3 sec it covers 21m

Then in 1 sec

21/3 = 7m/s

In 10sec

= 7 × 10

= 70m

Hence Verified