# Thread: Velocity/Acceleration Problem Help

1. ## Velocity/Acceleration Problem Help

"A particle moves along the x-axis so that its velocity at time T is given by:

v(T)= -(T+1)sin(T^2/2)

At T=0 the particle is at position x=1.

a) Find the acceleration of the particle at time T=2. Is the speed of the particle increasing at T=2? Why or why not?

b) Find all times T in the open interval 0<T<3 when the particle changes direction. Justify your answer. "

So I know to get the acceleration you have to derive the velocity function. This derivation consists of product quotient and chain rule.

So I got

V1(x)= Sin(t^2/2)*(-1)-(t+1)(cos(t^2/2))*(2*2t/4)

Then you would just substitute in 2 for t to find the acceleration at t=2 right? Is this the correct derivative of the velocity function, and how would you go about part b?

Thank you!

2. Originally Posted by roxy6
"A particle moves along the x-axis so that its velocity at time T is given by:

v(T)= -(T+1)sin(T^2/2)

At T=0 the particle is at position x=1.

a) Find the acceleration of the particle at time T=2. Is the speed of the particle increasing at T=2? Why or why not?

b) Find all times T in the open interval 0<T<3 when the particle changes direction. Justify your answer. "

So I know to get the acceleration you have to derive the velocity function. This derivation consists of product quotient and chain rule.

So I got

V1(x)= Sin(t^2/2)*(-1)-(t+1)(cos(t^2/2))*(2*2t/4)

Then you would just substitute in 2 for t to find the acceleration at t=2 right? Is this the correct derivative of the velocity function, and how would you go about part b?

Thank you!
$v(t) = -(t+1)\sin\left(\frac{t^2}{2}\right)$

$a(t) = -(t+1)\cos\left(\frac{t^2}{2}\right) \cdot t - \sin\left(\frac{t^2}{2}\right)$

calculate $a(2)$

for the (b) part, set $v(t) = 0$ and solve for t in the given interval. check the velocity on both sides of your solution to make sure it changes sign.

3. Ok so for part A I did:

a(2)= -(2+1)cos(2^2/2)*2-sin(2^2/2)

a(2)= -4.083.

The speed of the particle is decreasing because the acceleration is negative. Correct?

And for part B I did:

V(t) -(t+1)sin(t^2/2)=0
V(0)= -(0+1)sin(0^2/2)=0
V(1)= -(1+1)sin(1^2/2)= -1.438
V(2) = -(2+2)sin(2^2/2)= -3.637
V(3)= -(3+2)sin(3^2/2) = 4.887

The particle changes direction when t=3 because the sign goes from negative to positive.

Are these answers correct?

4. Originally Posted by roxy6
Ok so for part A I did:

a(2)= -(2+1)cos(2^2/2)*2-sin(2^2/2)

a(2)= -4.083. no

The speed of the particle is decreasing because the acceleration is negative. Correct? half correct ... your calculation and reasoning is wrong.

And for part B I did:

V(t) -(t+1)sin(t^2/2)=0
V(0)= -(0+1)sin(0^2/2)=0
V(1)= -(1+1)sin(1^2/2)= -1.438
V(2) = -(2+2)sin(2^2/2)= -3.637
V(3)= -(3+2)sin(3^2/2) = 4.887

The particle changes direction when t=3 because the sign goes from negative to positive.

no ... the particle changes direction when $\textcolor{red}{\frac{t^2}{2} = \pi}$

...

5. I recalculated and got a(2)= 1.5875

So the particle is speeding up because the sign is positive right?

I am having a problem with part b and have been trying it for 2 hours now. Could you show me how to do it step by step?

6. Originally Posted by roxy6
I recalculated and got a(2)= 1.5875

So the particle is speeding up because the sign is positive right? no, again.

the particle is slowing down because velocity and acceleration have opposite signs at t = 2. a(2) > 0 , v(2) < 0

I am having a problem with part b and have been trying it for 2 hours now. Could you show me how to do it step by step?
set $v(t) = 0$ ...

$-(t+1)\sin\left(\frac{t^2}{2}\right) = 0$

set each factor equal to 0 ...

$t+1 = 0$ at $t = -1$, but that value of $t$ is not in the interval $0 < t < 3$

$\sin\left(\frac{t^2}{2}\right) = 0$

$\frac{t^2}{2} = 0$ , $\pi$ , $2\pi$ , ...

the only value that will get $t$ in the desired interval $0 < t < 3$ is

$\frac{t^2}{2} = \pi$

$t^2 = 2\pi$

$t = \sqrt{2\pi} \approx 2.507$

and the velocity changes sign there also.