Question

Did I derive \(v = u+ at\) correctly?

Answer 1

Start with the definition of acceleration.\[ a = {v - u \over t}\]Multiply \(t\) to both sides.\[at = v - u\]Add \(u\) to both sides.\[v = u + at\]

Answer 2

yes....but this is valid for constant acceleration...so why u r using the formula a=(v-u)/t

Answer 3

Is there another way to derive?

Answer 4

yes ..u can use differential calculus

Answer 5

use dv/dt=a

Answer 6

and do d integration with proper limits, u'll get d same result

Answer 7

Oh yes!

Answer 8

Can you do it for me? @akash123

Answer 9

How do I start?

Answer 10

dv/dt=a

Answer 11

Yes, but how to continue?

Answer 12

What should I integrate?

Answer 13

dv=a dt

Answer 14

Yes, I follow. Then?

Answer 15

put integration sign on both sides

Answer 16

\[\int dv = \int adt\]

Answer 17

at t=0, v=u and t=t, v=u

Answer 18

yes...put the lower and upper limit on both sides

Answer 19

I'm new to the applications of calculus. :(

Answer 20

and pull out the acceleration a outside since it's a constant

Answer 21

ok...wait

Answer 22

\[\int\limits_{u}^{v}dv=a \int\limits_{0}^{t}dt\]

Answer 23

now it's fine?

Answer 24

Yes, very fine.

Answer 25

now do the integration

Answer 26

and put the limits

Answer 27

Ok, I got how to do this. Thank you!