Question

A body which uniformly accelerated and have an initial velocity, travels 200 cm in the first 2 seconds and 220 cm in the next 5 seconds. Calculate the velocity of the body at the end of the 7th second from the start, assuming that the acceleration throughout the journey is uniform.

Kindly solve this question with clear an full explanation.

Answer 1

Answer :-

Velocity of the body at the end of the 7th second from the start is 4 cm/s .

Explanation :-

We have :-

→ First distance = 200 cm

→ First time = 2 sec

→ Second distance = 220 cm

→ Second time = 5 seconds .

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For first 2 seconds :-

By using the 2nd equation of motion :-

s = ut + ½at²

⇒ 200 = u(2) + ½ × a × (2)²

⇒ 200 = 2u + ½ × a × 4

⇒ 200 = 2u + 2a ---(1)

For next 5 seconds :-

• Time = (2 + 5) = 7 seconds

• Distance = (200 + 220) = 420 cm

Again by the 2nd equation of motion :-

⇒ 420 = u(7) + ½ × a × (7)²

⇒ 420 = 7u + 24.5a ---(2)

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Equation (1) can be written as :-

⇒ 200 = 2(u + a)

⇒ 100 = u + a

⇒ u = (100 - a)

Substituting value of "a" in eq.2, we get :-

⇒ 420 = 7(100 - a) + 24.5a

⇒ 420 = 700 - 7a + 24.5a

⇒ -280 = 17.5a

⇒ a = -280/17.5

⇒ a = -16 cm/s²

Now, putting value of "a" in equation (1) :-

⇒ 200 = 2(u - 16)

⇒ 100 = u - 16

⇒ u = 116 cm/s

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We have got the Initial velocity and acceleration of the body. Finally we can calculate required velocity of the body by using the 1st equation of motion.

v = u + at

⇒v = 116 + (-16)7

⇒ v = 116 - 112

⇒ v = 4 cm/s

Answer 2

Given :-

A body which uniformly accelerated and have an initial velocity, travels 200 cm in the first 2 seconds and 220 cm in the next 5 seconds.

To Find :-

Calculate the velocity of the body at the end of the 7th second from the start, assuming that the acceleration throughout the journey is uniform.

Solution :-

Let

initial velocity = u

Acceleration = a

◼ I n c a s e 1

Distance = 200 cm

Time = 2 sec

s = ut + 1/2 at²

200 = u(2) + 1/2 × a × (2)²

200 = 2u + 1/2 × 4a

200 = 2u + 2a

Dividing both side by 2

200/2 = 2u + 2a/2

100 = u + a

100 - a = u (i)

◼ I n c a s e 2

Distance = 220 + 200 = 420 cm

Time = 2 + 5 = 7 sec

s = ut + 1/2 at²

420 = u(7) + 1/2 × a × (7)²

420 = 7u + a/2 × 49

420 = 7u + 24.5a

Putting value of u from 1

420 = 7(100 - a) + 24.5a

420 = 700 - 7a + 24.5a

420 = 700 + 17.5a

420 - 700 = 17.5a

-280 = 17.5a

-280/17.5 = a

-2800/175 = a

-16 = a

Now, Using 1

100 - (-16) = u

100 + 16 = u

116 = u

Now

v = u + at

v = 116 + (-16)(7)

v = 116 + (-112)

v = 116 - 112

v = 4 cm/s

∴ Acceleration = -16 cm/s² and Initial velocity = 116 cm/s and final velocity = 4 cm/s