Physics, asked by sultanmahmood197, 11 months ago

if the initial and final velocities of a moving body are 30 cm/sec and 3.70 meter/sec respectively then the distance covered in 5 seconds will be ?

Answers

Answered by joeashish2
3

Answer:

The answer is 14.43m/s

Explanation:

The initial velocity  is 30cm/sec = 0.3m/s

The final velocity is 3.7m/s

Time  = 5s

Therefore, the acceleration is v-u/t

                   =    3.7 - 0.3/5 = 0.68

Therefore the distance covered =  s=ut+1/2at^2

                                                  s = 0.3*5 + 1/2 *0.68*25

                                                    s = 14.34 metres.

HOPE THIS HELPS YOU...

Answered by VishalSharma01
51

Answer:

Explanation:

Solution,

Here, We have

Initial velocity, u = 30 cm/s = 0.3 m/s

Final velocity, v = 3.70 m/s

Time taken, t = 5 seconds

To Find,

Distance traveled, s = ?

At 1st we have to find the acceleration,

According to the 1st equation of motion,

We know that,

v = u + at

So, putting all the values, we get

v = u + at

⇒ 3.70 = 0.3 + a × 5

⇒ 3.70 - 0.3 = 5a

⇒ 3.4 = 5a

⇒ 3.4/5 = a

⇒ a = 0.68 m/s²

Here, the acceleration is 0.68 m/s².

Now, we will find the distance traveled,

According to the 3rd equation of motion,

We know that,

v² - u² = 2as

So, putting all the values, again we get

v² - u² = 2as

⇒ (3.70)² - (0.3)² = 2 × 0.68 × s

⇒ 13.69 - 0.09 = 1.36s

⇒ 13.6 = 1.36s

⇒ 13.6/1.36 = s

⇒ s = 10 m.

Hence, the distance covered by the body is 10 m.

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