Thread: Calculus, motion problem

I have a test tomorrow, and I would really appreciate any help on the following problems. Thanks a lot!!!!

1. A particle moves on a straight line so that its velocity at time t is given by v = 4s, where s is its distance from the origin. If s = 3 when t = 0, then when t = 1/2, s = ...
2. A particle moves along a line so that its acceleration, a , at time t is a = -t^2. If the particle is at the origin when t = 0 and 3 units to the right of the origin when t = 1, then its velocity at t = 0 is...

Once again, thank you for any help!!!

Originally Posted by turtle

I have a test tomorrow, and I would really appreciate any help on the following problems. Thanks a lot!!!!

1. A particle moves on a straight line so that its velocity at time t is given by v = 4s, where s is its distance from the origin. If s = 3 when t = 0, then when t = 1/2, s = ...
2. A particle moves along a line so that its acceleration, a , at time t is a = -t^2. If the particle is at the origin when t = 0 and 3 units to the right of the origin when t = 1, then its velocity at t = 0 is...

Once again, thank you for any help!!!

1. v = 4s
v = ds/dt so
ds/dt = 4s
ds/(4s) = dt ... integrating both sides we get:
INT(1/(4s))ds = INT(1)dt
1/4*ln|s| = t + C
ln|s| = 4t + 4C ... we'll assume 4C is a constant, and write it as C for simplicity, and we can assume s is positive
ln(s) = 4t + C
s = C*e^(4t) ... now, we know s(0) = 3
3 = C*e^0 --> C = 3
s = 3e^(4t) ... now plug in t = 1/2
s = 3e^2

2. a = -t^2
dv/dt = -t^2
dv = (-t^2)dt ... integrate both sides
INT(1)dv = INT(-t^2)dt
v = -1/3*t^3 + C ... we need s(t) so,
ds/dt = -1/3*t^3 + C
ds = (-1/3*t^3 + C)dt ... integrate both sides
INT(1)ds = INT(-1/3*t^3 + C)dt
s = -1/12*t^4 + Ct + D ... we know s(0) = 0, so,
0 = D
s = -1/12*t^4 + Ct ... we know s(1) = 3, so,
3 = -1/12*(1) + C --> C = 13/12 ... we can plug this back in the v(t)
v = -1/3*t^3 + 13/12 ... now plug in t = 0
v = 13/12

I went through that kind of fast, so if you need me to explain any step, I can.