Question

The position of a particle for t > 0 is given by ????⃗ (????)=(m.

(a) What is the velocity as a function of time?
(b) What is the acceleration as a function of time?
(c) What is the particle’s velocity at t = 2.0 s?
(d) What is its speed at t = 1.0 s and t = 3.0 s?
(e) What is the average velocity between t = 1.0 s and t = 2.0 s?

Answer 1

(a)
`v(t) = (6.0ti - 21.0t^(2)j - 5.0) m/s`

(b)
`a(t) = (6.0i - 42.0tj) m/s^2`

(c)
`v(2.0) = (12.0i - 84.0j - 5.0) m/s`

(d)
`speed (t=1.0 s) = \sqrt{6^(2)+21^(2)+5^(2)}` = 22.41 m/s;
`speed (t= 3.0 s) = \sqrt{18^(2)+189^(2)+5^(2)}` = 189.92 m/s

(e) average velocity = (9i-21j-5) m/s

Step-by-step explanation:

For the given problem:

`r(t) = (3.0t^(2)i - 7.0t^(3)j - 5.0t - 2k) m`

(a) The velocity as a function of time: v(t) = dr/dt. Thus:

`v(t) = (6.0ti - 21.0t^(2)j - 5.0) m/s`

(b) The acceleration as a function of time: a(t) = dv/dt. Thus:

`a(t) = (6.0i - 42.0tj) m/s^2`

(c) The particle velocity at t= 2.0s. Using the equation in part (a);

`v(2.0) = (6.0*2.0i - 21.0(2.0)^(2)j - 5.0) m/s`

`v(2.0) = (12.0i - 84.0j - 5.0) m/s`

(d) Its speed at t=1.0s and t=3.0s

`speed = \sqrt{v_(x) ^(2)+v_(y) ^(2)}`

at t=1.0s

`speed = \sqrt{6^(2)+21^(2)+5^(2)}` = 22.41 m/s

at t= 3.0s

`speed = \sqrt{18^(2)+189^(2)+5^(2)}` = 189.92 m/s

(e) The average velocity between t = 1.0 s and t = 2.0 s. Using the equation for r(t).

`r(1.0) = (3.0i - 7.0j - 5.0 - 2k) m` m

`r(2.0) = (12.0i - 28.0j - 10.0 - 2k) m`

average velocity = Δr/Δt = (r2-r1)/(t2-t1) = (9i-21j-5) m/s