Piecewise Rectilinear Motion Integration

[LNH]

Homework Statement

Suppose that a particle moves along a line so that its velocity v at time t is given by
5t, if 0≤t<1
(6(t)^(1/2))-(1/t), if 1≤t
where t is in seconds and v is in centimeters per second (cm/s). Estimate the time(s) at which the particle is 4 cm from its starting position.

Homework Equations

s(t)=∫v(t)

The Attempt at a Solution

Took antiderivative of velocity piecewise function, resulting in:
s(t)=(5/2)t^2, if 0≤t<1
s(t)=4t^(3/2)−ln(|t|), if 1≤t
Solved first part of piecewise function for 4, resulting in t=((2)(10)^(1/2))/5)≈1.2649, which is out of the domain of this part of the piecewise function. I do not know how the second part of the piecewise function would be solved for 4.

 

[HallsofIvy]

You have not taken the anti-derivative correctly! You forgot the "constant of integration".

If v= ds/dt= 5t for t between 0 and 1, then s= (5/2)t^2+ C. If v= ds/dt= 6t- (1/t) for t> 1 then s= 3t^2- ln(t)+ D, where "C" and "D" are constants. Since the distance function is continuous (an object cannot "jump" from one location to another) the function must be contiuous: at t= 1, s(1)=(5/2)(1)^2+ C= 3(1)^2- ln(1)+ D so 5/2+ C= 3+ D. C= D+ 1/2.

Is the object able to go 4 cm from its starting point in the first second? If so we would have to have s(t)- s(0)= ((5/2)t^2+ C)- C= (5/2)t^2= 4. Then t^2= 8/5 which is larger than 1. No, the object cannot go 4 cm in the first second.

For t larger than 1 the object will be 4 cm away when, s(t)- s(0)= 3t^2- ln(t)+ D- C=3t^2- ln(t)+ D- (D+ 1/2)= 3t^2- ln(t)- 1/2= 4 or 3t^2- ln(t)= 9/2. That cannot be solved "analytically" and will have to be solved using a numerical method.

 

[LNH]

s= 3t^2- ln(t)+ D

How is the "3t^2" obtained? I expected "4t^(3/2)" because (1/2)+1=(3/2).

 

[LNH]

Bump. I thought the integration for a variable raised to a power is to add one to the exponent and divide the coefficient by this exponent. Thus, I do not understand how 3t^2 could be obtained.