plS help me in this problem..?

Question

A bus travels 400 m bet 2 stops. it starts from rest and accelerates at 1.5m/ssquared until it reaches a velocity of 9 m/s. the bus continues at this velocity and then deccelrates at 2 m/ssquared until it comes to a half. Find the total time for the journey.

Answer 1

Hey anygirl,
Here's how it is:

u = initial velocity,
v = final velocity,
a = acceleration/deceleration (note that when a body decelerates, its value is considered negative)
s1 = distance travlled during acceleration
t1 = time taken to cover the distance s1
s2 = distance travelled during deceleration.
t2 = time taken to cover the distance s2
s = distance travelled during constant velocity
Total distance travelled = 400 m

During acceleration------>
When a body starts at rest, u = 0
u = 0, a = 1.5m/s, v=9m/s
v = u + a*t
9 = 0 + 1.5*t1
=> t1 = 6 sec

Distance travelled during acceleration--->
v2 = u2 + 2*a*s
9*9 = 0 + 2*1.5*s1
81 = 3*s1
s1 = 27m

During deceleration------>
When a body comes to rest, v=0
u = 9, a = -2m/s, v=0m/s
v = u + a*t
0 = 9 - 2*t2
9 = 2*t2
=> t2 = 4.5 sec

Distance travelled during deceleration---->
v2 = u2 + 2as
0 = 9*9 - 2*2*s2
0 = 81 - 4*s2
81 = 4*s2
s2 = 20.25m

Total distance travelled during acceleration and deceleration = 27 + 20.25 = 47.25m
However, we know that the total distance travelled is 400 m. This implies that during the remaining distance, i.e., 400 - 47.35 = 352.75m, the object was in constant velocity(neither accelerating, nor decelerating).

During constant velocity---->
s = 352.75m, u = 9m (remember, the object is still travelling at 9 m/s), a = 0 (constant acceleration),
s = u*t + 1/2*a*t2
352.75 = 9*t + 0
=> t = 39.194 sec

Total time taken = t1+t2+t
Total time taken = 6+4.5+39.194
=>Total time taken = 49.694 sec

Hope this helps.

Answer 2

initial acceleration:

vf = vi + a t

9 = 0 + 1.5 t ..... t = 6 s

and during acceleration, the bus traveled

x1 = xi + vi t + 1/2 a t^2

x1 = 0 + 0 + 1/2 (1.5) 36 = 27 m

Do the same thing for deceleration.

vf = vi + a t

0 = 9 + 2 t ..... t = 4.5 s

and during acceleration, the bus traveled

x3 = xi + vi t + 1/2 a t^2

x3 = 0 + 0 + 1/2 (2) 20 = 20 m during deceleration

Subtract the distance traveled during acceleration and deceleration from the total 400m. That gives 353 m.

You get the time for the middle section by dividing the remaining distance by 9. 353/9 = 39 seconds

Now you know the time during acceleration, during deceleration, and during the constant velocity middle period. Add them up and you have your answer.

39 + 6 + 4.5 = 49 seconds (I've rounded and you should figure out how many significant figures the answer should be).

Aloha

Answer 3

The bus accelerates for 6 seconds, traveling 27 meters during that time. The bus decelerates for 4.5 seconds, traveling 20.25 meters during that time. Over the remaining 352.75 meters, it is traveling at 9 meters per second, or 39.1944 seconds. Adding things up, we have 49.6944 seconds.

Answer 4

Well Golllllllllly....fries with that ?

Answer 5

49.69444444 seconds.