Question

A particle has a velocity of v→(t)=5.0ti^+t2j^−2.0t3k^m/s.

(a) What is the acceleration function?

(b) What is the acceleration vector at t = 2.0 s? Find its magnitude and direction.

Answer 1

a)
`a=5 i+2t j - 6\ t^2k`

b)
`a=(1)/(24.83)(5i+4j-24k)\ m/s^2`

Step-by-step explanation:

Given that

v(t) = 5 t i + t² j - 2 t³ k

We know that acceleration a is given as

`a=(dv)/(dt)`

`(dv)/(dt)=5 i+2t j - 6\ t^2k`

`a=5 i+2t j - 6\ t^2k`

Therefore the acceleration function a will be

`a=5 i+2t j - 6\ t^2k`

The acceleration at t = 2 s

a= 5 i + 2 x 2 j - 6 x 2² k m/s²

a=5 i + 4 j -24 k m/s²

The magnitude of the acceleration will be

`a=√(5^2+4^2+24^2)\ m/s^2`

a= 24.83 m/s²

The direction of the acceleration a is given as

`a=(1)/(24.83)(5i+4j-24k)\ m/s^2`

a)
`a=5 i+2t j - 6\ t^2k`

b)
`a=(1)/(24.83)(5i+4j-24k)\ m/s^2`