A bus starts from rest and accelerates at 1.5m/s^2 until it reaches a velocity of 9m/s. The bus continues at t

Question

A bus starts from rest and accelerates at 1.5m/s^2 until it reaches a velocity of 9m/s. The bus continues at this velocity and then decelerates at -2m/s^2 until it comes to a stop 400m from its starting point. How much time did the bus take to cover 400 m.

Also I would like to know how you guys solve these kind of problems. Is it like thinking or just plugging it into a formula?

Answer 1

agree with Leny G

Answer 2

ewwwwww. physics.

Answer 3

Honestly, a question like this is so simple that you just have to plug it into a formula. When you start getting friction and angles and stuff that's not just 1 dimensional, you do need to start thinking a bit.

(Assume no friction for this question)

-------------------------

Accelerating part (time):

Vo = 0m/s

Vf = 9m/s

a = 1.5m/s^2

t = ?

Vf = Vo + at

9 = 1.5t

t = 6s

Accelerating part (distance):

Vo = 0m/s

Vf = 9m/s

a = 1.5m/s^2

d = ?

Vf^2 = Vo^2 + 2ad

81 = 3d

d = 27m

--------------------------

Braking part (time):

Vo = 9m/s

Vf = 0m/s

a = -2m/s^2

t = ?

Vf = Vo + at

0 = 9 - 2t

-9 = -2t

t = 4.5s

Braking part (distance):

Vo = 9m/s

Vf = 0m/s

a = -2m/s^2

d = ?

Vf^2 = Vo^2 + 2ad

-81 = -4d

d = 20.25m

-------------------------

6s + 4.5s = 10.5s

Ugh, I forgot about the 400m LOL. OK, basically you know that the total time for accelerating and decelerating is 10.5 seconds. However, you don't know the time spent traveling at the constant velocity of 9m/s. You can find it though because of the 400m.

So ... now we know the distance as well covered in only braking and accelerating, which is 27 + 20.25 = 47.25m. Therefore, the distance covered traveling at the constant velocity of 9m/s is 352.75m (I got this number from 400 - 47.25)

Finding the time traveled while at the constant velocity:

V = d/t

9 = 352.75 / t

t = 39.19s

39.19 + 10.5 = 49.7s.

The answer is 49.7 seconds. (Which rounds to 50s)

I'm adding stuff into my answer now. Sorry for the confusion.

I hope I was able to help! Sometimes you just have to try and picture the situation in your head and then think of the values that actually make sense and which equations you need to use. Also, drawing a picture can REALLY help if you're stuck.

Answer 4

Like my Physics teacher says: Step 1 read it. Step 2 extract the information Step 3 read it again. Then plug in the info into your equations and solve it.

You sort of have to ignore the story once you have the info, just work with the numbers and equations. Use the story to help make sense of it and if it sounds right or not.

Answer 5

It's not simply plugging into formulas, you need to understand the concepts.

First stage of the travel, a = 1.5 m/s^2

acceleration means change in speedwith time. given a = 1.5 m/s^2 means the velocity increases by 1.5 m/s for each second.

Therefore, when v = 9 m/s, the car has travelled 6 seconds.

a = v/t

1.5 = 9/t

t = 9/1.5

t = 6 sec

the average velocity is

v=(v1+v2)/2 = d/t

d = (v1 +v2)t/2

d = (0+9)6/2

d = 54/2 = 57 m

Second stage of travel, v = 9 m/s (constant)

velocity means the distance changes by 9 m for each second of travel

v = d/t

but we don't know d or t, thus we can not yet solve the time for this stage

Third stage of travel, deceleration a = -2 m/s^2

this means the velocity decreases by 2m/s for each second of travel, for a complete stop,

a = v/t

-2 = (0-9)/t

t = -9/-2

t =4.5 sec

the average velocity is v

(v1+v2)/2 = d/t

d = (v1 +v2)t/2

d = (0+9)4.5/2

d = 20.25/2 = 10.125 m

the total distance = 400 m

thus the distance for the second stage is

400 m - 57 m - 10.125 m = 332.875 m

v = 9 = 332.875/t

t = 36.99 sec or 37 sec

the total time is

t = 6 + 37 + 4.5 = 47.5 sec

Answer 6

the answer is green. seriously