Magnitude of velocity and acceleration

[PeroK]

Wouldn't that make c=0 at the start point?

$c = d_0 = 0$ from the way the problem is set up.

 

[JustDerek]

$c = d_0 = 0$ from the way the problem is set up.

So I make $a=0 s=524 and V_0=5$ and try to solve t for this?

 

[PeroK]

So I make $a=0 s=524 and V_0=5$ and try to solve t for this?

Setting $a = 0$ was just a way to see that your formula was wrong. If you integrate correctly you get:

$d = \frac{at^3}{6} + v_0t$

Now, you have to put some numbers in $a = 0.06$ and $v_0 = 5$ and use whatever numerical method you prefer to solve for $t$.

 

[JustDerek]

Setting $a = 0$ was just a way to see that your formula was wrong. If you integrate correctly you get:

$d = \frac{at^3}{6} + v_0t$

Now, you have to put some numbers in $a = 0.06$ and $v_0 = 5$ and use whatever numerical method you prefer to solve for $t$.

Which is get $524=0.01t^3+5t$ which believe it or not I've had worked out for two days. What I don't get is how to isolate t from this

 

[PeroK]

Which is get $524=0.01t^3+5t$ which believe it or not I've had worked out for two days. What I don't get is how to isolate t from this

I would just do it numerically. There is a way for solving cubic equations. Try googling if you interested. But, I'd just hit it with a spreadsheet.

 

[JustDerek]

I would just do it numerically. There is a way for solving cubic equations. Try googling if you interested. But, I'd just hit it with a spreadsheet.

Never done that before but I'll look into it. Won't that give multiple values for t?

 

[PeroK]

Never done that before but I'll look into it. Won't that give multiple values for t?

I know almost nothing about numerical methods. There is something called the Newton-Rhapson method, which you could codify in a program or spreadsheet. But, you know, all you have to do is check increasing values of $t$ until you go past the point you want, then refine the increments until you've narrowed it down to the required number of decimal places.

It's going to be somewhere between 30-40 seconds I guess.

You should be able to see analytically that there is only one real solution:your cubic function is monotonic.

 

[JustDerek]

I know almost nothing about numerical methods. There is something called the Newton-Rhapson method, which you could codify in a program or spreadsheet. But, you know, all you have to do is check increasing values of $t$ until you go past the point you want, then refine the increments until you've narrowed it down to the required number of decimal places.

It's going to be somewhere between 30-40 seconds I guess.

You should be able to see analytically that there is only one real solution:your cubic function is monotonic.

Thanks for all your help. I'll see what I can do from here. You've been a saint I'd give you +100 likes if I could

 

[PeroK]

Thanks for all your help. I'll see what I can do from here. You've been a saint I'd give you +100 likes if I could

One would do!

 

[Randy Beikmann]

When it is at point A then V_A=5m/s which increases at a rate of a=(0.06t)m/s².
Derek

Derek, I apologize. I missed the fact that the acceleration is not a constant! It looks like you got the help you needed though.

 

[JustDerek]

That's OK. The intention to help was appreciated