Question

Please help me out. :(

A body is moving at 40 m/s and its speed is decreasing uniformly at the rate of 5 m/s^2. Determine:

a) its speed at the end of 6 sec?
b) its average speed during 6 sec?
c) the distance passed over in 6 sec?

Answer 1

A)
First, times the acceleration by the time
\[5m/s^2*6=30\]
Take away the speed by this number
\[New Speed = OriginalSpeed - AboveNumber\]
\[40-30=10\]
The answer to A is 10

Answer 2

i dont get this.. please explain.. hehe. :)

Answer 3

Ok, first you know that his speed is decreasing at a rate of \[5ms^{-2}\]
And you know this goes on for 6 seconds, as said in the question. Therefore, if you times the rate by 6, you can get how much the speed has decreased, here \[30ms^{-1}\]
If you take away the total, here 40, from 30, you will get the current speed at 6 seconds, @itsmeeQUENNIE

Answer 4

Does that help? @itsmeeQUENNIE

Answer 5

ahhh.. yeah.. i got it. thank you for explaining it clearly. :)

Answer 6

No problem, now to the next questions :)

Answer 7

If, at any point, you are confused, just ask :)

Answer 8

i used the formula..
Average Velocity = (v + u) / 2

Answer 9

The equation for acceleration:
\[Acceleration=\frac{v^{2}+v^{1}}2\]
\[v^2=EndSpeed\]
\[v^1=BeginningSpeed\]
Sub in the values
\[Acceleration=\frac{50+10}2\]
\[Acceleration=30\]

Answer 10

I meant average speed*

Answer 11

3)
\[AverageDistance=AverageSpeed*Time\]
\[Distance=30*6\]
\[Distance=180m\]

Answer 12

That's it :)

Answer 13

Anything you don't understand, @itsmeeQUENNIE?

Answer 14

what is the answer in letter b?

Answer 15

Answer for B) is 30m/s

Answer 16

Using the rule you learn:
\[AverageVelocity=\frac{v+u}{2}\]

Answer 17

am, @Ahsome , why 50 ? why not 10 ? please explain. thankyou.

Answer 18

@itsmeeQUENNIE What do you mean 50 and not 10? I used both in that equation

Answer 19

ahm.. i mean.. why 50 is the final velocity?

Answer 20

Sorry, I mean 10+50

Answer 21

I put them in the wrong place ;)

Answer 22

So it would be:
\[AverageSpeed=\frac{10+50}{2}\]

Answer 23

why 50 is the initial? how did you solve that?

Answer 24

What am I doing. I meant 40 for the initial. My keyboard is stuck :(

Answer 25

So, everytime I use 50, just put 40.

Answer 26

@itsmeeQUENNIE Now do you understand?

Answer 27

so, what's the final answer for letter b? :)

Answer 28

To find Average Speed:
\[AverageSpeed=\frac{40+10}{2}\]
\[AverageSpeed=\frac{50}{2}\]
\[AverageSpeed=25ms^{-2}\]
To find Average Distance:
\[AverageDistance=AverageSpeed∗Time\]
\[Distance=25∗6\]
\[Distance=150m\]

Answer 29

in letter c, why did you use the answer in letter b? please explain it again. thankyou. :)

Answer 30

u=40m/s
a=-5m/s^2 (retardation)
(a) t=6sec
v=u+at
v=40-5*6=40-30=10m/s
(b)
average speed
\[=\frac{ u+v }{ 2 }=\frac{ 40+10 }{ 2 }=25 m/s\]
(c)
\[s=ut+\frac{ 1 }{ 2 }a t^2=40*6+\frac{ 1 }{ 2 }*(-5)*6^2=240-90=150 m\]
or distance=average speed *time=25*6=150m

Answer 31

Sure :)
They asked you the distance they made in that six seconds. Remember the equation: \(Distance=Speed*Time\). Since the average speed is the overall speed he had in those 6 seconds, we use that.

We don't use 50, because he didn't travle for \(50ms^{-1}\) for 6 seconds. He started at that speed, and slowly went down to \(10ms^{-1}\) at a rate of \(5ms^{-2}\). Instead of trying to use all that info, we just used the average speed.

Kinda like how the average height is around 5Feet, 7Inches. Not many people ARE that height, but for mathematical reasons we can use it to cover all.

Answer 32

@surjithayer I was WONDERING what you were typing there ;)

Answer 33

There are two ways, either using the \(s=ut+12at^{2}\), as shown by @surjithayer. Or you can use the \(Distance=AverageSpeed*Time\)

Answer 34

Hopefully that helps you, @itsmeeQUENNIE

Answer 35

I guess.. i'll take the solution of @surjithayer ..
i understand it easily without question on my mind. :)

sorry @Ahsome . :( really really sorry.

Answer 36

we can also use
\[v^2-u^2=2as,\]
\[\left( 10 \right)^2-\left( 40 \right)^2=2*(-5) s\]
100-1600=-10s
-1500=-10s
\[s=\frac{ -1500 }{ -10 }=150m\]

Answer 37

@itsmeeQUENNIE It's OK :) Whatever floats your boat :)

Answer 38

@surjithayer .. ah.. but im okay with your 1st solution..

Answer 39

@Ahsome .. thank you so much for understanding.. :)

Answer 40

No Problem

Answer 41

you are welcome.
i have only given you the various solutions for knowledge .

Answer 42

thankyou. :) ^_^