Question

A body is moving with uniform acceleration. Its velocity after
5seconds is 25m/s and after 8 seconds is 34 m/s . How much distance
will it cover in 10 seconds after its start.

Answer 1

$\huge \underline \green{\sf{ Answer :- }}$

V = u + at

After 5s

25 = u + 5a ——(1)

After 8s

34 = u + 8a ——(2)

Subtract equation (1) from equation (2)

34 - 25 = u + 8a - u - 5a

9 = 3a

a = 3 m/s^2

Substitute a = 3m/s^2 in equation (1)

25 = u + 5*3

u = 10 m/s

Distance travelled in 12th second

S = u + a(n - 1/2)

S = 10 + 3*(12 - 1/2)

S = 44.5m

Distance travelled in 12th second is 44.5m

Answer 2

Answer

• Distance travelled will be 2500 m

Explanation

Given

• Acceleration of a body is uniform
• Its velocity after 5 seconds is 25 m/s and after 8 seconds is 34 m/s

To Find

• Distance covered after 10 seconds

Solution

[So here we may use the first Equation of motion and with its help form two linear equations, so will these equations find the acceleration of the body. So then the distance covered will be given by the second Equation of motion]

✭ After 5 Seconds

➝ v = u+at

• v = 25 m/s
• Time = 5 sec

➝ 25 = u+5a -eq(1)

✭ After 8 Seconds

➝ v = u+at

• v = 34 m/s
• Time = 8 sec

➝ 34 = u+8a -eq(2)

✭ Subtracting eq(1) from eq(2)

➝ 34-25 = (u+8a) - (u+5a)

➝ 9 = u+8a-u-5a

➝ 9 = 3a

➝ 9/3 = a

➝ Acceleration = 3 m/s²

✭ Initial Velocity

➝ 25 = u+5×3

[From eq(1)]

➝ 25 = u+15

➝ 25-15 = u

➝ Initial Velocity = 10 m/s

✭ Distance covered in 10 seconds

➝ s = ut+½at²

➝ s = 10 × 10 + ½ × 3 × 10²

➝ s = 1000 + ½ × 3 × 1000

➝ s = 1000+3 × 500

➝ s = 1000+1500

➝ Distance Travelled = 2500 m