a body is coasting to a stop and the only force acting on it is a resistance

**yoman360** · March 26th 2010, 04:27 PM

A body is coasting to a stop and the only force acting on it is a resistance proportional to its speed, according to the equation:
$\frac{ds}{dt}=v_f=v_0e^{(\frac{-kt}{m})}$ ; s(0)=0, where $v_0$ is the body's initial velocity (in m/s), $v_f$ is its final velocity, m is its mass, k is a constant, and t is time.

(a) If a body with mass m = 50 kg and k = 1.5 kg/sec initially has a velocity of 30 m/s, how long, to the nearest second, will it take to slow to 1 m/s?

(b) How far, to the nearest 10 meters, will the body coast during the time takes to slow from 30 m/s to 1 m/s?

(c) If the body coasts from 30 m/s to a stop, how far will it coast?

**hollywood** · March 26th 2010, 06:33 PM

You can get (a) by using the given differential equation and solving for t.

Since $v_0\text{, }k,\text{ and }m$ are constants, you have $\frac{ds}{dt}$ equal to a function of t, so you only have to take the integral of both sides to get s as a function of t.

If you don't know how to integrate, try setting $s=c_1e^{c_2t}+C$ , take the derivative and choose values of $c_1\text{ and }c_2$ to make the equation work. You'll have to use the initial condition s(0)=0 to get C.

The answer to (c) is that the question is invalid - if you look at the differential equation, you see that $v_f$ never reaches zero, so it never stops. But you can take the limit as t goes to infinity of the equation you found in (b) to get the result.

Post again in this thread if you're still having trouble.

**skeeter** · March 26th 2010, 06:39 PM

Originally Posted by yoman360

A body is coasting to a stop and the only force acting on it is a resistance proportional to its speed, according to the equation:
$\frac{ds}{dt}=v_f=v_0e^{(\frac{-kt}{m})}$ ; s(0)=0, where $v_0$ is the body's initial velocity (in m/s), $v_f$ is its final velocity, m is its mass, k is a constant, and t is time.

(a) If a body with mass m = 50 kg and k = 1.5 kg/sec initially has a velocity of 30 m/s, how long, to the nearest second, will it take to slow to 1 m/s?

(b) How far, to the nearest 10 meters, will the body coast during the time takes to slow from 30 m/s to 1 m/s?

(c) If the body coasts from 30 m/s to a stop, how far will it coast?

(a) solve the equation $v(t) = v_0 e^{-\frac{kt}{m}}$ for $t$ , then sub in your given values to calculate $t$

(b) using $t$ found in part (a) as the upper limit of integration ...

$d = \int_0^t v(t) \, dt$

(c) $d = \lim_{b \to \infty} \int_0^b v(t) \, dt$

**yoman360** · March 26th 2010, 08:23 PM

Originally Posted by skeeter

(a) solve the equation $v(t) = v_0 e^{-\frac{kt}{m}}$ for $t$ , then sub in your given values to calculate $t$

(b) using $t$ found in part (a) as the upper limit of integration ...

$d = \int_0^t v(t) \, dt$

(c) $d = \lim_{b \to \infty} \int_0^b v(t) \, dt$

I got t=0 but i think i did something wrong

**hollywood** · March 27th 2010, 12:16 AM

I assume you mean for (a). You should be setting $v_0=30\text{, }k=1.5\text{, }m=50\text{, and }v_f=1$ . If you still get t=0, please show us your work and one of us will show you where you went wrong.

- Hollywood

**yoman360** · March 27th 2010, 08:57 AM

Originally Posted by hollywood

I assume you mean for (a). You should be setting $v_0=30\text{, }k=1.5\text{, }m=50\text{, and }v_f=1$ . If you still get t=0, please show us your work and one of us will show you where you went wrong.

- Hollywood

$v_f=v_0e^{(\frac{-kt}{m})}$

$1=30e^{(\frac{-1.5t}{50})}$

when making this post i realized the mistake i made

now i get

$t=\frac{-50}{1.5}*ln(\frac{1}{30})$

(A) t≈ 133 seconds

hopefully thats correct

**skeeter** · March 27th 2010, 09:10 AM

Originally Posted by yoman360

$v_f=v_0e^{(\frac{-kt}{m})}$

$1=30e^{(\frac{-1.5t}{50})}$

when making this post i realized the mistake i made

now i get

$t=\frac{-50}{1.5}*ln(\frac{1}{30})$

hopefully thats correct

correct ... that is the value of t such that v(t) = 1

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