Question

Is my solution right?
Problem: Two cars start to constantly accelerate from one point to the same direction. After time t_1 the distance between them is l. How much time t_2 would need to pass, so that the distance between the cars would be equal to 3l?

Answer 1

\(a = \frac{v}{t}\)
\(v = \frac{s}{t}\)
so, \(a = \frac{s}{t^2}; s = at^2\)
Let's say that a_2 > a_1;
Then we make a system of equations:
\(a_2 t_1^2 - a_1 t_1^2 = l\) -> \(t_1^2(a_2 - a_1) = l\)
\(a_2 t_2^2 - a_1 t_2^2 = 3l\) -> \(t_2^2(a_2 - a_1) = 3l\)

then we just put the l from the first equation into the second and solve for t_2 to get:
\(t_2 = \sqrt{3t_1^2} = t_1\sqrt{3}\)

Is that the right way to solve this problem? @ganeshie8

Answer 2

\(\color{green}{\Huge \checkmark }\)

Answer 3

Your formula ` v = s/t ` doesn't look correct.

Aren't you wrongly assuming that the velocity is constant ?

Answer 4

@ganeshie8 you are right, however, how could I get distance from only acceleration and time?

Answer 5

We use the given info that acceleration is constant

Answer 6

How good are you with calculus - integration and differentiation ?

Answer 7

Only basics.

Answer 8

Basics will do.
What's the relation between acceleration and velocity ?

Answer 9

a = v/t

Answer 10

That works only when acceleration is constant. Mayble let me ask this first :

What's the relation between velocity and displacement ?

Answer 11

Maybe dv/dt?

Answer 12

Though that's exactly the same...

Answer 13

Yes,

a = dv/dt

Answer 14

Next, what's the relation between velocity and displacement ?

Answer 15

v = s/t

Answer 16

Again, that works only when velocity is constant. Think..

Answer 17

v ds/dt

Answer 18

You mean v = ds/dt

Answer 19

yes.

Answer 20

Imagine you're driving and your displacement from your home is given by the equation

s = 3t^2

What is the function that gives your velocity at ANY point in time ?

Answer 21

Well, it would be F(t) = 3t^2

Answer 22

s = 3t^2 is your displacement from the home.

How can your velocity be same as displacement ?

Answer 23

Ups. Sorry. I still haven't waken up.
s/3t^2 = v

Answer 24

Haha I see you need some good coffee :)

s = 3t^2 implies s/3t^2 = 1, not s/3t^2 = v

Answer 25

Oh. Yes, I must have a cup. :D
so, it would be the 3t^2/t = v; 3t = v

Answer 26

Wrong. As I've said earlier, there is no reason to assume that velocity is constant.

v = s/t works only when velocity is constant.

Answer 27

But how would we find the d of 3t^2?

Answer 28

Yes, we must use the relation between velocity and displacement :

v = ds/dt

Answer 29

(0 - 3t^2)/(0 -t) ?

Answer 30

For now memorize this formula :
\[\large \dfrac{d}{dt} (t^n) = n*t^{n-1}\]

Answer 31

\[\large \dfrac{d}{dt} (t^2) = ?\]

Answer 32

2t

Answer 33

\[\large \dfrac{d}{dt} (3t^2) = ?\]

Answer 34

6t

Answer 35

good

Answer 36

So, if I give you a function for displacement, you can find the corresponding function for velocity ?

Answer 37

I give you `displacement`

you can give me `velocity `

Answer 38

good so far.

what if i gave you the velocity function and asked you to find the corresponding displacement function ?

Answer 39

ds/dt = v
ds = dt * v

Answer 40

I want "s", not "ds"

Answer 41

I have no idea how to cancel out the d ;)

Answer 42

Take the previous function itself :

v = 6t

I want you to recover the corresponding displacement function.

(assume your starting velocity is 0)

Answer 43

I know you can't do it yet.. let's take a quick look at a more simpler problem

Answer 44

Alright :)

Answer 45

Suppose you're cycling at a constant velocity of 6 m/s.

How much distance would you be travelling in 5 seconds ?

Answer 46

v*s = 30 meters

Answer 47

Let's work it in another way (this method is going to make you love calculus)

Answer 48

Let's first plot the velocity function : v(t) = 6