please help. a student is running at constant velocity of +5.0m/s in an effort to catch a bus?
parked at a station. when he is 10.0 m away from the bus, the bus pulls away with a constant acceleration of + 1.2 m/s². how much time will it take the student to catch the bus if he keeps running with the same velocity?
solutions and answer please :)
- Anonymous9 years agoFavorite Answer
Velocity = +5 m/s
Distance = 10 m
Acceleration of Bus = +1.2 m/s²
Let the student catch the bus after time 't'
In time 't', the bus travels -->
h = ut + 1/2at² = 0 + 1/2(1.2)(t²) = 0.6 t²
In time t, the student travels -->
distance = speed x time = 5t
Total distance traveled by Boy = (Initial 10 meters) + (Additional Distance the bus travels)
5t = 10 + 0.6t²
t = 5 sec or 3.33 sec
Well you have two values of time and it seems neither of them can be rejected in any ways.
- helooLv 49 years ago
constant velocity for the student so 5 = d+10 / t
d= 0.5 a t^2
sub back in
0.6 t^2 +10 /t =5
0.6t^2 +10 =5t
3t^2 - 25t +50 = 0
(3t - 10)(t - 5) = 0
t = 5