Question

a car travels along a straight line at a constant speedof 60.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. the average velocity of the entire trip is 30.0 mi/h (a) what is the constant speed with which the car moved during the second distance d?

Answer 1

i would assume that the second speed is slower than 30?

Answer 2

and how do i set this up @rajathsbhat

Answer 3

is it (60 + x)/2 = 30??

Answer 4

since distance is the same

Answer 5

*thinking*

Answer 6

It's 20 mph.

Answer 7

why???

Answer 8

t1=d/60
t2=d/x

30=2d/t1+t2
substitute & simplify

Answer 9

2d/ti?

Answer 10

2d/t1?

Answer 11

ohhh the formula...2d/t1 + t2

Answer 12

yes.

Answer 13

formula is \[\color{red}{V_{av.}=\frac{2V_1V_2}{V_1+V_2}.}\]

Answer 14

\[\large 30 = \frac{2d}{\frac{(60+x)d}{60x}}\] that right?

Answer 15

yep.

Answer 16

then \[\large 30 = \frac{60x}{60+x}\]

Answer 17

\[180 = 30x\]

xshould be 60? o.O

Answer 18

1800

Answer 19

oh wait i missed 2

Answer 20

and that.

Answer 21

\[1800 + 30x = 120x\]

\[1800 = 90x\]

x should be 20?

Answer 22

the constant speed is 20 m/h. here's how you do it.
it travels d distance with 60m/h, so time taken t1= d/60
it travels another d distance with a velocity say v, time taken t2= d/v
hence total time taken t = t1+t2 = d/60 + d/v
now avg velocity = total distance / total time
= 2d/ t
= 2d/ (d/60 + d/v)
= 120v/ (v +60)
It's given that avg velocity = 30m/h= 120v /(v+60)
if you solve it, v = 20m/h

i hope this is correct

Answer 23

tada.

Answer 24

great!!1 thanks

Answer 25

yw :)

Answer 26

uhmm woops forgot there was follow up question

Answer 27

(b) suppose the second distance d were traveled in the opposite direction what is the average velocity for the trip? 0 right?

Answer 28

here the second trip is in the opposite direction of the first trip and we're asked the vector velocity.
you can close your eyes and say that the answer is zero. Because when we find the avg velocity we divide avg displacement by total time (formula in part (a) ). because the displacement is zero ( d+ (-d) = 0 ), the avg velocity of obviously zero.

Answer 29

very right @lgbasallote gr8 work

Answer 30

yes, lgba

Answer 31

yayyy

Answer 32

:) yahoo....

Answer 33

(C) what is the average speed

average speed is 2d/t1 + t2 right?

Answer 34

here we have been asked the speed, so it doesn't matter in which direction we move. We have to focus on the magnitude alone. And the magnitudes have been given the same as that in part (a). hence the answer for avg speed = 30 m/h (which has been already provided in the question itself).

Answer 35

yes

Answer 36

If a particle moves a distance at a speed v1 & again same distance with v2
then the formula is: -\[\color{red}{V_{av.}=\frac{2V_1V_2}{V_1+V_2}.}\]If a particle moves a distance in 2 equal intervals of time at different speeds v1 & v2 respectively then the formula is: -\[\color{red}{V_{av.}=\frac{V_1+V_2}{2}.}\]Where "Vav." is average velocity.

Answer 37

wow, @mathslover how did you type so much before i could type yes?

Answer 38

lolz .... i was prepared for that .. i had that question in my book

Answer 39

hahaha..

Answer 40

and also i am preparing for this topic ..... all lgba's question i had done just 1-2 days ago .:)

Answer 41

so t1 = d/60

t2 = d/-20 <---because it also said that to return trip was the same constant speed as found in part (a)

so then \[\frac{2d}{\frac{d}{60} - \frac{d}{20}}\] that right?

Answer 42

No, for speed, there's no + or - stuff. It's + always.

Answer 43

nope ... it is not fully right

Answer 44

yes since speed is scalar quantity

Answer 45

it does not include direction it only includes magnitude no direction

Answer 46

t2=d/20

Answer 47

oh so it's d/60 + d/20?

Answer 48

right t2 = \(\frac{d}{20}\) right

Answer 49

yep.

Answer 50

\[\color{green}{\text{V=speed}}\]\[\color{blue}{{\overrightarrow V}=\text{Velocity} }\]

Answer 51

lolz @maheshmeghwal9 thankss ... i deleted that as it was taking so much space

Answer 52

okay so summary \[\huge V_{ave} = \frac{2d}{t_1 + t_2}\]

right?

Answer 53

yw:)

Answer 54

yes, lgba.

Answer 55

right

Answer 56

wait doesnt this mean the average speed for round trip is the same as the average velocity of (a)???

Answer 57

yeah, it does.

Answer 58

yes.. it is

Answer 59

hmm great thanks

Answer 60

that's the point.Speed=don't care about the direction