Question

The position of a particle for t > 0 is given by ?⃗(?) = (3.0 ? ?̂ − 7.0 ? ?̂ − 5.0 ? ?) 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? Assume all variables and constants are in SI units.

Answer 1

Given position

r(t) = (3.0t i – 7.0t 1 - 5.0t k) m

(a)

Velocity is the time derivative of position

(t)

ü(t) = (3.0t î– 7.0t j - 5.0+ k) m

ū(t) = (3.0 i - 7.0 j -5.0 k) m/s

(b)

The acceleration is time derivative of velocity

a(t) = "(t)

ā(t) = (3.0 i – 7.0 9 – 5.0 k)

a(t) = 0

(c)

We calculated velocity as a function of time part (a)

ū(t) = (3.0 i – 70 j - 5.0 k) m/s

It is constant in time. At time t=2.0s

ū(t = 2.0s) = (3.0 i - 7.0-5.0 k) m/s

(d)

The speed is the magnitude of velocity

ū= 13.0 i – 7.0 - 5.0 k

Jū1 = 3.02 +(-7.0)2 i +(-5.0)2 m/s

u = 9.11 m/s

The speed does not depend on time. So, the speeds at time t=1.0s and t=3.0s are equal.

(e)

The position at time t=1.0s is

r(t) = (3.0t i – 7.0t 1 - 5.0t k) m

(1.0 s) = (3.0 x (1.0s) i - 7.0 x (1.0s) 1 - 5.0 x (1.0s) k) m

(1.0 s) = (3.0 i - 7.0 - 5.0 k) m

The position at time t=2.0s is

(2.0 s) = (3.0 x (2.0s) i - 7.0 x (2.0s) 1 - 5.0 x (2.0s) k) m

(2.0 s) = (6.0 i – 14.0 j - 10.0 k) m

The average velocity of the particle between time t=1.0s and t=2.0s is

Ūavg = F(2.0 s) – F(1.0 s) 2.0s – 1.0s

aq =. (6.0 î - 14.0 j - 10.0 k) m - (3.0 î – 7.0 j - 5.0 k) m) 2.0s – 1.0s

Vavy = (3.0 i – 70 j - 5.0 k) m 1.0s

Tavg = (3.0 i - 7.0 j -5.0 k) m/s