Homework Help: Acceleration, Velocity and Position (calculus)

1. The problem statement, all variables and given/known data

The acceleration of a particle is directly proportional to time, t. At t = 0, the velocity of that particle, v, = 16 in./sec. Knowing that v = 15 in/sec and x = 20in when t = 1, determine the velocity, acceleration and position at x = 7.

2. Relevant equations

3. The attempt at a solution

I took
[tex]\frac{dv}{dt} = kt, so \: v = \frac{k}{2}t^2 + C[/tex]

Solving for C:
[tex]16 = \frac{k}{2}(0)^2 + C[/tex]
C = 16

Solve for k

[tex]1 = \frac{k}{2}(1)^2 + 16[/tex]

k = -30

Therefore:

[tex]v(t) = \-15*(t)^2 + 16[/tex]

This is already wrong; v(7) = -719 in/sec but the book states the answer is -33 in/sec

Any advice on what's gone wrong would be greatly appreciated; this is a revision question too...

Cheers,
Adrian

 

Ah, thankyou ehild; I failed to read the question once again... I will give it a shot. How is k = -2?

 

I found x(t) by integrating and finding C. I obtained this:

[tex]x(t)=16t-5t^3+9[/tex]

Then solving for t when x = 7

[tex]16t-5t^3+2 = 0[/tex]

Sorry, how does one solve this?

Cheers,
Adrian

 

Sorry, this is probably really stupid, but where/how does k = -2?

 

Ah! I put in a bad number...dodgy mistake indeed; Cheers, I will retry!

 

Got it!! Thanks so much, both of you! ;)