A particle moves according to a law of motion s = f(t), t 0, where t is measured

Question

A particle moves according to a law of motion s = f(t), t 0, where t is measured in seconds and s in feet. (If an answer does not exist, enter DNE.) f(t) = t3 8t2 + 22t

(a) Find the velocity at time t.

v(t) =  ft/s

(b) What is the velocity after 1 second?

v(1) =  ft/s

(c) When is the particle at rest?


(d) When is the particle moving in the positive direction?


(e) Find the total distance traveled during the first 6 seconds.
ft

(f) Draw a diagram like this figure to illustrate the motion of the particle.

(g) Find the acceleration at time t and after 1 second.

(h) Graph the position, velocity, and acceleration functions for

0 t 6.

(A graphing calculator is recommended.)

(i) When is the particle speeding up? (Enter your answer using interval notation.)


When is it slowing down? (Enter your answer using interval notation.)

Explanation / Answer

s = f(t)

f(t) = t3 8t2 + 22t

(a). Velocity at time t,

v(t) = f'(t)   (velocity is change in position with time)

v(t) = (3t2 - 16t + 22) ft/s

(b). Velocity after 1 sec.

v(1) = 3(1)2 - 16(1) + 22

= 3 - 16 + 22

= 25 - 16 = 9 ft/s

(c). Particle is at rest when velocity is zero.

v(t) = 0

3t2 - 16t + 22 = 0

Since the quadratic equation has no real roots,

The particle is never at rest. It is continuously moving.

(d). Particle will be moving in positive direction, when the value of v(t) will be positive,

v(t) = 3t2 - 16t + 22

As v(t) is equal to zero for no real value of t,

=> v(t) will always be positive (greater than 0)

Therefore, particle will always be moving in positive direction.

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