Argh!! Really stuck!! Please help

**yummyburgers** · April 9th 2009, 10:55 AM

Hi all, I'm having a few problems doing practical problems. I can solve some of them but I'm having problems with deaccelleration in particular.. Can you help me solve these problems?

The distance s travelled by a train in t seconds after it applies its brakes is given by

. How far has the train travelled before it comes to rest ?

I'm totally stumped by this one but I can do some acceleration questions, such as:

A missile is fired vertically upwards and its height h in metres after time t is given by

. Calculate the maximum height reached by the missile.

I know thw anwer to this is 1000.. But I'm lost with deaccelleration..

Can someone please help, thanks in advance.

**nzmathman** · April 9th 2009, 12:58 PM

If $s = 16t-0.4t^2$ then v (velocity) $= \frac{ds}{dt} = 16-0.8t$
$v=0$ when the train is at rest. So solve for t. Then plug this value of t in to the distance equation to find the total distance traveled.

$h=500+100t-5t^2$
To get v, differentiate. So $v = 100-10t$
You should know the equation for h represents the graph of a parabola. So for the maximum h the derivative, v, is 0. Solve $v=0$ , plug the t value into the equation for h and you will have your answer of 1000m.

**yummyburgers** · April 10th 2009, 03:47 AM

I've tried to find t but I don't know what it is?

I tried using 0.8 as t and using the formula as:

s= 16t - 0.4t to be squared
s= 16(0.8) - 0.4(0.8) to be squared but the answer I got was 12.8 and that's wrong...

Could you please tell me the answer and how you found it..? Thanks.

**mr fantastic** · April 10th 2009, 03:55 AM

Originally Posted by nzmathman

If $s = 16t-0.4t^2$ then v (velocity) $= \frac{ds}{dt} = 16-0.8t$

$v=0$ when the train is at rest. So solve for t. Then plug this value of t in to the distance equation to find the total distance traveled.
[snip]

Originally Posted by yummyburgers

I've tried to find t but I don't know what it is?

I tried using 0.8 as t and using the formula as:

s= 16t - 0.4t to be squared
s= 16(0.8) - 0.4(0.8) to be squared but the answer I got was 12.8 and that's wrong...

Could you please tell me the answer and how you found it..? Thanks.

When you were instructed to solve for $t$ , that meant solve $v = 0$ for $t$ , that is, solve $\frac{ds}{dt} = 0$ for $t$ , that is, solve $16 - 0.8t = 0$ for $t$ .

**yummyburgers** · April 10th 2009, 04:22 AM

So that would mean ds/dt = 16t -0.4t to be squared..?

I tried subtracting 16 - 0.4 = 15.6 and 16 - 0.8 = 15.2but those answers are wrong?

**mr fantastic** · April 10th 2009, 04:27 AM

Originally Posted by yummyburgers

So that would mean ds/dt = 16t -0.4t to be squared..?

I tried subtracting 16 - 0.4 = 15.6 and 16 - 0.8 = 15.2but those answers are wrong?

Originally Posted by mr fantastic

When you were instructed to solve for $t$ , that meant solve $v = 0$ for $t$ , that is, solve $\frac{ds}{dt} = 0$ for $t$ , that is, solve
16 - 0.8t = 0 for $t$ .

Can you solve the red equation for t? Can you substitute that value of t into $s = 16 t - 0.4 t^2$ ?

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