<p>At time t=0 a jogger is running at a velocity of 300 meters per minute. The jogger is slowing down with a negative acceleration that is directly proportional to time t. This brings the jogger to a stop in 10 minutes</p>
<p>a)write an expression for the velocity of the jogger at time t</p>
<p>b)what is the total distance traveled by the jogger in that 10 minute interval?</p>
<p>I never did a problem like this but here goes.
V = v' + a(t) and v=0, v' = 300 and a = kt
so 0 = 300 + kt^2 and k = -8.33<em>10^-4 m/s
v(t) = 300 + (-8.33</em>10^-4)t^2;
s = 300<em>(10min</em>60s) + 1/2(k*t) * t^2 = 90036 m
is it even logical, it would be but I messed up and didnt change the initial 300 to m/s. Just change that and you should get an answer that makes sense.</p>
<p>i dont understand how you did this problem (how did you solve for k) and what do u have to change to get a sensible answer?</p>
<p>can anyone help?</p>
<p>My answer is wrong, I think really wrong. You are going to have to integrate the acceleration to find something, I am not really sure. My class just did constant acceleration. Ill check it up in my book now and post again later.</p>
<p>okay thanks matt</p>
<p>I'm assuming that it is constant acceleration. Then simply:</p>
<p>V = Vo + at
0 = 300 + 10a
a = -30 m/min
then</p>
<p>V = 300 - 30t
and then integrate</p>
<p>d = 300t - 15t^2 + do(which is 0)
then solve for an answer of 1500 m</p>
<p>I think that is right.</p>
<p>acceleration is NOT constant
"jogger is slowing down with a negative acceleration that is directly proportional to time t. "</p>
<p>how do you do this, anybody please help</p>
<p>OK with variable acceleration then</p>
<p>a = kt
V = Vo + .5kt^2
0 = 300 + .5k*10^2
k = - 6 m/min^4
then</p>
<p>V = 300 - 3t^2
d = 300t - t^3 (worked out nicely)
then d = 2000 m, which now should be the right answer.</p>
<p>It is essentially asking you to write an equation with the acceleration... so to find that:</p>
<p>Vf = Vi + a<em>t, where Vf = final velocity and Vi = initial velocity
0 = 300 + a</em>10
a = -30 m/s^2</p>
<p>So an expression could be: f(t) = 300 - 30*t, where t is in minutes</p>
<p>To find the distance traveled, use:
d = Vi<em>t + 1/2 (a</em>t^2)
d = 300(10)+1/2(-30*100)
d = 1500 meters</p>