Question

g A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.03t3 (a) Find the velocity at time t (in ft/s). v(t) = 0.04t^3−0.09t^2 Correct: Your answer is correct. (b) What is the velocity after 1 second(s)? v(1) = -0.05 Correct: Your answer is correct. ft/s (c) When is the particle at rest? t = 0 Correct: Your answer is correct

Answer 1

Answer:

Explanation:

If a particle move with time and expressed according to the formula:

f(t) = 0.01t⁴ − 0.03t³

a) Velocity is the change in motion of the particle with respect to time and it is expressed as;

$v(t) =\frac{d(f(t))}{dt}$

$v(t) = 4(0.01)t^{4-1} - 3(0.03)t^{3-1}\\v(t) = 0.04t^3 - 0.09t^2$

Hence the velocity of the particle at time t is $v(t) = 0.04t^3 - 0.09t^2$

b) To calculate the velocity after 1 second, we will substitute t = 1 into the function v(t) in (a) as shown:

$v(t) = 0.04t^3 - 0.09t^2\\v(1) = 0.04(1)^3 - 0.09(1)^2\\v(t) = 0.04 - 0.09\\v(t) = -0.05$

Hence the velocity after 1second is -0.05

c) The particle is at rest when when the time is zero.

Initially, the body is not moving and the time during this time is 0. Hence the particle is at rest when t = 0second