I have trouble solving this calculus problem! An object travels in a straight tunnel with acceleration. 4 m/s2. The tunnel is 60 meters long, and takes 5 seconds in order to reach the final. Find the initial speed and instant speed when you have 20 meters left to go:. What I did was a (t) = 4 v (t) v = 4t + C1 (5) = 4 (5) + C1 = 0 C1 = -20 (this is the initial velocity I think) But my problem is that resolution second part, I have no idea what I have tried to do is: v (t) = 4t-20 s (t) = 2t ^ 2-20t + C2 do not know what to do after much appreciated! Any explanation be.
if the tunnel is 60 m long and it takes 5 seconds you can use the simple division of knowing that the average of the V is 12 miles per hour, which is exactly speed of 2.5 seconds in the tunnel. therefore 12 – (2.5 * 4) = V initial = 2 m / s, now that the initial V and the initial position, you can connect to the shall make its distance (derivitave of the equation of the speed) with C1 = 2, then you solve for D = 40 I would say that 40 = 2t ^ 2 + 2t + 0 (initial position) to solve t = -5 t = 4 because it is using the time, negative numbers can not be used to test t = 4: V medium for 4 to 5 seconds should be 20 m / s (because there are just 20 meters left to go in the tunnel at t = 4) with its acceleration be 4 m / s / s, this means that final v (t = 5) should be 22 m / s, yvyt = 4 should be 18 4 (acceleration) * 4 (time) + 2 (initial velocity) = 18 m / s 4 (acceleration) * 5 (time) + 2 (speed initial) = 22 m / s 22 m / s is the same speed you get from the equation of the drift speed let me know if that helps