Solution:
Distance, Velocity - time functions are linked easily through derivation and integration:
Distance - time function → derivation → Velocity - time function
Velocity - time function → derivation → Acceleration - time function
(and vice versa)
Let's assume we have a distance - time function:
[tex]s(t) = 4t^{2} - 2t +7[/tex]
where s is measured in feet and t in seconds.
a) To find velocity at time t, we simply derivate the distance - time function:
[tex]\frac{ds}{dt} = v(t) = 8t - 2[/tex]
b) To find velocity at t-3, we simply substitute 3 in the velocity - time function:
[tex]v (t) = 8t -2\\v(3) = 8(3) -2\\v(3) = 22 \ ft/sec[/tex]
c) A particle will be at rest when it's velocity is zero. Thus, we substitute v = 0 in the velocity - time function:
[tex]v (t ) = 8t -2\\8t -2 = 0\\8t = 2\\\\t = \frac{2}{8}\\\\t= \frac{1}{4} seconds[/tex]
Hence, at time t = 1/4 seconds, the object will be at rest.
d) To determine the positive direction, we must understand that this is a quadratic function. Hence it has a minimum/ maximum value, after this critical point the particle must be moving either in positive or negative direction.
Hence, we find this critical point. A critical point of any function is it's derivative equalled to zero.
The derivative of distance - time function is a velocity - time function. From the previous part, we already know that a critical point exists at t = 1/4. Now, we substitute, t = 1/4, in the distance - time function to find the other co-ordinate:
[tex]s (t) = 4t^{2} - 2t +7\\s(\frac{1}{4}) = 4(\frac{1}{4})^{2} - 2(\frac{1}{4})+7\\\\s(\frac{1}{4}) = 4(\frac{1}{16}) - 2(\frac{1}{4})+7\\\\s(\frac{1}{4}) = \frac{1}{4} - \frac{2}{4}+\frac{28}{4}\\\\s(\frac{1}{4}) = \frac{27}{4} \\\\[/tex]
The function will be positive after [tex](\frac{1}{4}, \frac{27}{4})[/tex]
e) The total distance travelled in first 8 seconds can be determined by substituting t = 8 in distance - time function:
[tex]s(t) = 4t^{2} - 2t+7\\\\s(8) = 4(8)^{2} - 2(8)+7\\\\s(8) = 4 (64) - 2 (8) +7\\\\s(8) = 247 feet[/tex]
