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# 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 :)

thank you.

### 2 Answers

- 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

But,

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.

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- helooLv 49 years ago
constant velocity for the student so 5 = d+10 / t

for bus:

s =d

u=0

v=

a=1.2

t

d= 0.5 a t^2

d=0.6 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

t=3.33