Question

If s = 2t3 – 3t2 + 2, find the position, velocity and acceleration of a particle at the end of 2 second. s is measured in meter and t is in second.

Answer 1

Explanation:

Correct option is

B

18m/s

correct answer is B.

a=3t

2

+2t+2

dv=adt

v=

3

3t

3

+

2

2t

2

+2t

v=t

3

+t

2

+2t

at t=0,u=2

att=2

v=t

3

+t

2

+2t+2

=8+4+2×2+2

=8+4+4+2

=18m/s

Answer 2

Answer: At end of 2 seconds, particles position is 6 m , its velocity is 12m/s and its acceleration is 18 m/s²

Explanation:

s = 2t³ - 3t² + 2 ( in metres)

At t = 2,

s = 2(2)³ - 3(2)² + 2

s = 16 - 12 + 2

s = 6

Position at end of 2 second is 6m.

Velocity is defined as rate of change of position

⇒ $v = \frac{ds}{dt}$

⇒ $v = \frac{d(2t^{3}-3t^{2} + 2)}{dt}$

⇒ v  =  6t² - 6t

At t = 2,

v = 6(2)² - 6(2)

v = 12 m/s

Velocity at end of 2 second is 12m/s

Acceleration is defined as rate of change of velocity

⇒ $a = \frac{dv}{dt}$

⇒ $a = \frac{d(6t^{2}-6t) }{dt}$

⇒ a = 12t - 6

At t = 2,

a = 12(2) - 6

a = 18 m/s²

Acceleration at end of 2 second is 18 m/s²

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