How Do You Calculate the Speed of a Particle Given Its Position Function c(t)?

  • Thread starter Thread starter Tubig
  • Start date Start date
  • Tags Tags
    Particle Speed
Tubig
Messages
2
Reaction score
0
1. Find the speed (As a function of t) of a particle whose position at time t seconds is c(t)=(sint+t,cost+t)
2. speed = [x'(t)^2 + y'(t)^2]^1/2
3.
x'(t) = cost + 1
y'(t)= -sint + 1

speed = [(cos t + 1)^2 +(-sint + 1)^2]^1/2
=[cos^2 t+ 2cost + 1 +sin^2 t - 2sint + 1]^1/2
=[cos^2 t + sin^2 t + 2 cost - 2sint +2]^1/2
=[1 + 2(cost - sint + 1)]^1/2
-Thanks
 
Physics news on Phys.org
Or (2*(cos(t)-sin(t))+3)^(1/2). That looks ok to me.
 
No, Dick. It would be (2cos(t)- 2sin(t))+3)^(1/2). The "3" would not be multiplied by 2.
 
HallsofIvy said:
No, Dick. It would be (2cos(t)- 2sin(t))+3)^(1/2). The "3" would not be multiplied by 2.

No, Halls. There's an extra set of parentheses around the cos(t)-sin(t). When you cut and pasted you only deleted one of them.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top