How Far Does a Bus Travel After Accelerating and Decelerating?

[exemplarysam]

there is this problem that i really need help on

<b>a bus accelerates at 1.5m/s(squared) from rest for 12s. Itthen travels at constant velocity for 25s, after which it slows to a stop with an acceleration of -1.5m/s(squared).
(a) How far does the bus travel?
(b) What is its average velocity?

 

[Vodka]

i won't give it away completely, but you'll be needing these equations [not in that particular order]

d = v\t

a = (v' - v) / t

d = [(v' - v)/2]t

v'^2 = v^2 + 2ad

isolation and substition will be involved :)

 

[M98Ranger]

</b>

(a) To find the distance traveled by the bus, we can use the formula d= ut + 1/2at^2, where d is the distance, u is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is 0 m/s, the acceleration is 1.5m/s(squared) and the time is 12s. Plugging these values into the formula, we get d= (0)(12) + 1/2(1.5)(12)^2 = 108m. Therefore, the bus travels 108 meters in the first 12 seconds.

Next, the bus travels at a constant velocity for 25 seconds, which means its acceleration is 0 m/s(squared). Using the formula d= ut + 1/2at^2, where a= 0, we get d= (12)(25) + 1/2(0)(25)^2 = 300m. Therefore, the bus travels an additional 300 meters in the next 25 seconds.

Finally, the bus slows down with an acceleration of -1.5m/s(squared) until it comes to a stop. Using the same formula, d= ut + 1/2at^2, where a= -1.5, we get d= (12)(25) + 1/2(-1.5)(25)^2 = 150m. Therefore, the bus travels an additional 150 meters before coming to a stop.

The total distance traveled by the bus is 108 + 300 + 150 = 558 meters.

(b) To find the average velocity of the bus, we can use the formula v= d/t, where v is the average velocity, d is the distance traveled and t is the total time taken. In this case, the total time taken is 12 + 25 + 25 = 62 seconds. Substituting the values, we get v= 558/62 = 9 m/s. Therefore, the average velocity of the bus is 9 meters per second.