Answer to Question #9907 in Mechanics | Relativity for Franconise Charles
seconds, the truck moved:
60.0 = 0.5*2.20 * t^2, so t = sqrt(2*60.0 / 2.20) =
approx. 7.385 s.
b)How far the automobile behind the truck
Let that distance was n meters.
When they met, automobile
60.0 + n = 3.50 * t^2 = 3.50 * 27.2727 = 95.4545 m, so
n = 95.4545 - 60.0 = 35.45 m.
c)What is the velocity of each when they are next to
Truck: Vt = at * t = 2.20 * 7.385 = 16.25 m/s.
Va = aa * t = 3.50 * 7.385 = 25.85 m/s.
0.5*t^2 = 0.5*7.385^2 = 27.27
how did you get 27.2727 ?
Dear Chris, find fixed answer in the attachment.
These are horribly wrong.
Please write a new question
Shouldn't part a be 60 = (1/2)2.2t^2? Missing a one-half there.
Can you p[lease solve this. I hardly can't answer it. Pleaseee
A car and a truck start from rest at the same instant with the car initially at some distance behind the truck. The truck has a constant acceleration of 3.4 m/s^2. The car overtakes the truck when the truck has moved a distance of 40.0 m.
a. a)How much time does it take the car to overtake the truck? b)How far the car behind the truck initially? c)What is the velocity of each when they are next to each other?