Question

A particle moves along a straight line .The position of the particle at time t is given by x=40+12t-t^3 . Calculate the average velocity in first 2 s

Answer 1

ANSWER:

• Average Velocity = 8 m/s.

GIVEN:

• The position of the particle at time t is given by x = 40 + 12t - t³.

• Time = 2 s.

TO FIND:

• The average velocity.

EXPLANATION:

$\boxed{ \bold{Avg \ velocity = \dfrac{Total \ displacement}{Time \ Taken}}}$

Let us find the diaplacement.

Let us call the initial point as O. [ when t = 0 ]

When t = 0

x = 40 + 12(0) - 0³

x = 40 + 0 + 0

x = 40 m

Initial position is x = 40

Let us call the final point as A. [ when t = 2 ]

When t = 2

x = 40 + 12(2) - 2³

x = 40 + 24 - 8

x = 40 + 16

x = 56 m

Final position is x = 56 m

Total displacement = Final position - Initial position

Total displacement = 56 - 40

Total displacement = 16 m.

Total time = 2 s

$\tt Avg \ velocity = \dfrac{16}{2}$

$\tt Avg \ velocity = 8 \ m {s}^{ - 1}$

HENCE THE AVERAGE VELOCITY = 8 m/s