A truck on a straight road starts from rest and accelerates at 2.8 m/s^2 until it reaches a speed of 16 m/s. Then the truck travels for 16s at constant speed until the brakes are applied,stopping the truck in a uniform manner in an additional 3.6 s.
How long is the truck in motion?
Answer in units of s
What is the average velocity of the truck for the motion described?
Answer in units of m/s
So I've done the first part already which is 25.31 s, but when i try to find the average velocity for part two. I keep getting it wrong...can anyone show me the process on how to correctly do the second one?
V = Vo + a*t = 16 m/s.
0 + 2.8*t = 16
t1 = 5.71 s. to reach 16 m/s.
d1 = 0.5a*t^2 = 0.5*2.8*5.71^2 = 45.65 m
d2 = V*t2 = 16 * 16 = 256 m.
t1+t2+t3 = 5.71 + 16 + 3.6 = 25.31 s in
motion.
V = Vo + a*t = 0
16 + a*3.6 = 0
3.6a = -16
a = -4.44 m/s^2.
d3 = (V^2-V0^2)/2a = (0-(16^2)/-8.88 =
28.8 m. To stop.
Vavg = (d1+d2+D3)/(t1+t2+t3) =
(45.65+256+28.8)/25.31 = 13.06 m/s.
To find the average velocity of the truck during the second part of its motion, we need to determine the distance traveled during that time.
In the second part, the truck travels at a constant speed for 16 seconds. Since the speed is constant, we can use the formula average velocity = total distance / total time.
To find the total distance, we can multiply the constant speed of the truck (16 m/s) by the time it travels at that speed (16 s):
Total distance = constant speed × time
Total distance = 16 m/s × 16 s
Total distance = 256 meters
Now we can calculate the average velocity:
Average velocity = Total distance / Total time
Average velocity = 256 meters / (16 + 3.6) seconds
Average velocity = 256 meters / 19.6 seconds
Average velocity ≈ 13.06 m/s
So the average velocity of the truck during the second part of its motion is approximately 13.06 m/s.