pleaaaaaaaaaase i need help with this physics problem!?

Question

A record of travel along a straight path is as follows:
1. Start from rest with constant acceleration of 2.05 m/s2 for 20.0 s
2. Constant velocity for the next 2.70 min
3. Constant negative acceleration 8.77 m/s2 for 4.73 s
(a) What was the total displacement for the complete trip?
m
(b) What were the average speeds for legs 1, 2, and 3 of the trip as well as for the complete trip?
m/s
m/s
m/s
m/s

Answer 1

Ok...lets do some calculations to find out what we need to know for the 2 questions.

1. If the acceleration is constant and for a set time you'll be able to measure your velocity for the 20 s.

Vfinal = u (initial velocity) + (A * T) (Acceleration * Time)

Vfinal = 0 m/s + (2.05 m/s^2 * 20.0 s)

Vfinal = 2.05 m/s^2 * 20.0 s

Vfinal = 2.05*20.0 m*s/s^2

The units will cancel out like a variable...

Vfinal = 41 m/s


2. Now we're travelling at our final velocity with no acceleration for 2.70 minutes, we have to use dimensional analysis to convert this to seconds to find our distance.

2.7 min. x 60 sec / 1 min = x

(2.7 * 60 min*sec) / min = x

162 sec = x

Now we know it was travelling at the final velocity in leg 1 for 162 seconds.

Now to calculate our distance it's simply D = VT

Distance = velocity * time

D = 41 m/s * 162 s

D = 6642 m*s/s (seconds cancel out)

D = 6642 m



3. Now negative acceleration is basically deceleration or coming to a stop.

We'll use the same formula we had before, we'll simply change the sign of the acceleration value and figure out our displacement from there.

Vfinal = u + at

Vfinal = 41 m/s + (-8.77 m/s^2 * 4.73s)

Now we use 41 m/s because that was our final velocity during the initial acceleration, and we were not accelerating faster than that, but held it when we got there.

Vfinal = 41 m/s - 41 m/s

Vfinal = 0 m/s.


Now that we have the legs figured out, we can find the answers to the 2 questions.




a. Total displacement can be calculated as follows.

D = Leg 1 + Leg 2 + Leg 3

Leg 1 and 3 displacement requires a different formula because there was acceleration involved, we already know the displacement on leg 2 because we calculated it earlier, it was 6642 m.

s = ut + 1/2at^2 (displacement = initial velocity * time + 1/2 * acceleration * time^2)

Leg 1 displacement

s1 = (0 * 20.0 s) + ((.5 * 2.05 m/s^2) * (20.0s)^2)

s1 = 1.025 m/s^2 * 400 s^2

s1 = 410 m



Leg 3 displacement

s3 = ut + 1/2at^2

s3 = (41 m/s * 4.73s) + ((.5 * -8.7 m/s^2) * ((4.73s)^2))

s3 = 193.93 m + (-4.35 m/s^2 * 22.37s^2)

s3 = 193.93 m - 97.31m

s3 = 96.62 m


Now that we know the lengths of the three legs of the trip, we can add them up for a total displacement.

s final = s1+ s2 + s3

s final = 410 m + 6642 m + 96.62 m

s final = 7149 m Final answer.


b. Ok, for this one we can calculate it quite easily since the average speed is basically average velocity. So it's the difference in velocity over time. This equation for the first and 3rd leg is only for constant acceleration.

1. 1st leg.

V avg = ((u + v) / 2)

V avg = ((0 + 41 m/s) / 2)

V avg = 20.5 m/s

V avg = 20.5 m/s Final answer.


2. 2nd leg = 41 m/s Final answer.

The reason being is that it's a constant velocity, and unchanging so it'll be the average velocity.


3. 3rd leg

V avg = ((u + v) / 2)

V avg = ((41 m/s + 0 m/s) / 2)

V avg = 20.5 m/s Final answer.


4. Whole trip

V avg = (Leg 1 + Leg 2 + Leg 3) / 3

V avg = (20.5 m/s + 41 m/s + 20.5 m/s) / 3

V avg = 82 m/s / 3

V avg = 27.33 m/s Final answer.


I realize it's a long answer, but I hope it helps you with your understanding of physics.

Answer 2

no