finding average speed?

Question

A record of travel along a straight path is as follows.
1. Start from rest with constant acceleration of 2.71 m/s^2 for 15.1 s.
2. Constant velocity for the next 3.47 min.
3. Constant negative acceleration -9.68 m/s^2 for 3.39 s.

a) What was the average speed for leg 3 of the trip in m/s?
b) What was the average speed for the complete trip in m/s??

I tried this problem and I keep getting it wrong, I have no idea what I am doing wrong! Please help! THANKS!

Answer 1

a. To find the average speed V during leg 3 you need to know the distance D and time T since V = D/T.

We know D = v0t + (1/2)aT^2
v0 = velocity at start of leg 3 which is the same as that at the end of leg 1 since velocity is constant during leg 2.
a = acceleration during leg 3 = -9.68 m/s^2
T = time length of leg 3 = 3.39 seconds

We need to determine v0 and this is done as mentioned above from leg 1. Use:
v = at where a = 2.71 m/s^2 and t =15.1 seconds
v = 2.71*15.1 = 40.921 m/s = v0

D = (40.921)(3.39) - (1/2)(9.68)(3.39)^2
D = 83.1 meters

V = D/T = 83.1/3.39 = 24.51 m/s

b. Need distance from leg 1 and leg 2. We already have all the times and these sum to:
T = 15.1 + 3.47(60) + 3.39 seconds
T = 225.69 seconds

Distance leg 1: D1 = (1/2)at^2 = (1/2)2.71(15.1)^2
D1 = 308.95 meters

Distance leg 2: D2 = (velocity at end leg1)x(time)
D2 = 40.921 *(3.47*60)
D2 = 8519.7522 meters

Distance leg 3: determine in part a.
D3 = 83.1

D = D1 + D2 + D3 = 308.95 + 8519.7522 + 83.1
D = 8911.8022 meters

V = D/T = 8911.8022 / 225.69
V 30.487 m/s

Answer 2

First, figure out Leg 1. Find the distance traveled (d1), and the final velocity (v1), because you'll need those later. Use these formulas:

d1 = v0(t) + at²/2
or (since initial velocity v0 is zero):
d1 = at²/2

and:

v1 = v0 + at
= at

(you are given "a" and "t", so just plug them in).

2. Figure out distance & final speed in leg 2, using similar formulas:

d2 = v1(t) + at²/2
or (since a=0 for this leg):
d2 = v1(t)

also:
v2 = v1 (since the speed doesn't change in this leg)

3. Figure out the distance traveled in leg 3:

d3 = v2(t) + at²/2

Now you have all the numbers you need:

a) What was the average speed for leg 3 of the trip in m/s?

avg speed = distance/time = d3/(3.39 s)

b) What was the average speed for the complete trip in m/s?

avg speed = distance/time = (d1+d2+d3)/(15.1s + (3.47min)(60s/min) + 3.39s)

Answer 3

D1 = Distance travelled during acceleration
u = initial velocity = 0
v = velocity after 15.1 s
a = accelaration = 2.71 m/s^2

v = u + at

v = (2.71)*(15.1) = 40.921 m/s

D1 = u*t + (0.5)*a*(t^2)
D1 = (0.5)*(2.71)*(15.1)*(15.1)
D1 = 308.95m

============================================

D2 = Distance travelled during constant speed

= speed*time
= 40.921*3.47*60 (since time is in minutes)
= 8519.75m

============================================

D3 = Distance travelled during decelaration

Intial veloci of final leg: v = 40.921 m/s
Final velocity v1
Final Leg time t1 = 3.39s
Deceleration a1 = 9.68 m/s^2

v1 = v - a1*t1
v1 = 40.921 - 9.68*3.39
= 8.1 m/s

D3 = v*t - (0.5)*a1*(t1^2)

D3 = 40.921*3.39 - (0.5)*9.68*3.39*3.39
= 83.1m

============================================
Average Speed for leg 3
= 83.1/3.39 = 24.51m/s

============================================
Average Speed for whole trip
= (D1+D2+D3)/(total time)

Total distance = (308.95+8519.75+83.1)
= 8911.8 m

Total Time = (15.1+(3.47*60) + 3.39)
= 226.69

Average Speed for whole trip
=8911.8/226.69
=39.31 m/s

Answer 4

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