Question

On a distant planet a body, starting from rest falls freely for 1.5 sec. Calculate the distance travelled by the body assuming it to be possessing a uniform acceleration for 20 m/s² .

Answer 1

Given:

• initial velocity = 0 m/s
• time = 1.5 seconds

• acceleration = 20 m/s^2

To find:

distance = ?

We know Newton's second kinematic equation i.e, s = ut + 1/2 × at^2

Solution:

(just put the values)

--> s = 0 × 1.5 + 1/2 × 20 × (1.5)^2

--> s = 0 + 2 × 2.25

--> s = 4.5 meters

Hence, we get the required distance as 4.5 m

Hope this helped you dear...

Answer 2

Answer :

›»› The Distance travelled by body = 22.5 m

Given :

• Initial velocity of body (u) = 0 m/s
• Time taken by body (t) = 1.5 sec
• Acceleration of if body (a) = 20 m/s²

To Calculate :

• Distance travelled by body (s) = ?

How To Calculate ?

❏ As per it is given in the question that a distant planet a body, starting from rest falls freely so its initial velocity represented by 'u' will be 0 m/s (because starting from rest),Time taken is 1.5 sec, and Acceleration is also given which 20 m/s²$.$

❏ To calculate the Distance travelled by we'll use the second equation of motion which says s = ut + ½ at²

Where,

• s is the Distance travelled in m.
• u is the Initial velocity in m/s.
• t is the Time taken in second.
• a is the Acceleration in m/s².

Required Calculation :

† From second equation of motion

⇒ s = ut + 1/2 at²

⇒ s = 0 × 1.5 + 1/2 × 20 × 1.5²

⇒ s = 0 + 1/2 × 20 × 2.25

⇒ s = 1/2 × 20 × 2.25

⇒ s = 1 × 10 × 2.25

⇒ s = 10 × 2.25

⇒ s = 22.5

║Hence, the Distance travelled by body is 22.5 m.║