Question

Q 45 A particle having initial velocity 3 m/s is
moving with constant acceleration 2m/s2 for t
= 4 seconds. Calculate the displacement of the
particle during the last second.

Answer 1

Answer:

10 metres

Explanation:

Given :

• Initial velocity of the particle = u = 3 m/s

• Acceleration of the particle = a = 2 m/s²

• Time taken = t = 4 seconds

To find :

• Displacement in the 4th second of motion

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

S (4) = 3+2/2(2×4-1)

S (4) = 3+1(7)

S (4) = 3+7

S (4) = 10 metres

The distance covered in the 4th second of motion is equal to 10 metres

Answer 2

Given that:

• Initial velocity (u) = 3 m/s

• Acceleration(a) = 2 m/s^2

• Time = 4 seconds.

To Find:

• Displacement in the 4th seconds.

Formula Used:

• Sn = u +a(2n-1)/2

where,

Sn is the distance travelled in nth second

u is the initial velocity

a is the acceleration

n is the time

SOLUTION:

Using the above formula, we have

Sn = (3) + 2(2*4-1)/2

Sn = 3+7

Sn = 10 m

So, the displacement in the 4th second is 10 m.

Hope this helps