Calculus, particle motion?

Question

A particle moves according to a law of motion given by f(t) = t^2 - 16t +1 (with t>=0) and t is measured in seconds and f(t) in feet. Find the total distance travelled in the first 15 seconds.

Those are the answers that I got, but the options are "The total distance travelled D(15) = 120.5, 113, 117, or 114.5 feet". Not 14

Answer 1

f(0) = 1
f(15) = 15^2 - 26*15 + 1 = -14

What needs to be found is what happens in between.

Maximum displacement of the particle will occur at f'(t) = 0.
f'(t) = 2t - 16
When f'(t) = 0,
2t - 16 = 0
2t = 16
t = 8

Particle position at that time will be
f(8) = 8^2 - 16*8 + 1 = -63

The particle travels from 1 to -63, then from -63 to -14.
The total distance traveled (disregarding vector directionality) is thus:
|(1-(-63))| + |(-63 -(-14))| = 64 + 49 = 113.

Answer 2

The question asks for the distance travelled, not the displacement. If it requested displacement, then I would evaluate the forrmula straight with t = 15

Ok. Evaluate the formula at t = 0 to get the starting point

f(t=0) = 1

Evaluate again at t =15 to get the end point.

f(t=15) = 15^2 -16*15 + 1 = -14

Hence the distance travelled is one minus the other which equals 15 feet.

Answer 3

f(0) = 0^2 - 16×0 +1
=1

f(15) = 15^2 -16×15 +1
= 225 - 240 +1
= -14 feet

But since distances cannot be negative, we have to take the abosolute value of -14 feet which is |-14 feet| = 14 feet.

Finally find the difference.
f(15) -f(0)
=14 -1
=13 feet

Answer 4

For area C because the different guy responded A&B: enable the curve of C be defined as: C: x = t^3 ? 3t, y = t^2 ? 2t, ?10 ? t ? 10 to ensure that the graph to be transferring quickly down at a level, the spinoff of the graph could could be undefined. keep in mind that, for parametric curves: dy/dx = [dy/dt] / [dx/dt] utilising this formulation for the spinoff of the curve you'll discover that, to ensure that the spinoff of the curve to be undefined, the spinoff of x with respect to the parameter should be 0. So, differentiate x with respect to t: x = t^3 ? 3t dx/dt = 3t^2 - 3 And enable dx/dt = 0: 0 = 3t^2 - 3 t = ±a million we've 2 solutions at present, and both are on the period of t, yet because of the requirement of the region, we could make efficient that the particle is transferring down at this factor. ordinary approaches to do that is to seem on the spinoff of y with respect to t to work out no matter if that's detrimental at both of those values. dy/dt = 2t - 2 at the same time as t = -a million, dy/dt = -4 ... lowering. at the same time as t = a million, dy/dt = 0 ... The particle has stopped transferring. the purely fee at which the particle is transferring quickly down is at the same time as t = -4. hence: t = -a million ~~~

Answer 5

the distance traveled is the first derivative of the speed

f(t)=t^2-16t+1

f'(t)=d=2t-16

when t=15

d=2(15)-16
d=14 feet