Question

The acceleration "a" in m/s^2 of a particle is given by 3t^2+2t+2,where "t" is the time.If the partile starts out with velocity V=2m/s at t=0,then,the velocity at the end of 2 second is?

Answer 1

Do you know how to transform expression of acceleration to velocity?

Answer 2

no..

Answer 3

Do you have definitions of displacement, velocity, and acceleration?

Answer 4

yes

Answer 5

So, what is the definition of acceleration?

Answer 6

displacement=shortest distance between 2 initial points
velocity=rate of change of speed per unit time
acceleration=rate of change of velocity per unit time

Answer 7

"acceleration=rate of change of velocity per unit time"

so, mathematically, acceleration is obtained by derivation of velocity and velocity by the opposite transformation, i.e. integration.

Answer 8

i guessed that correct in my mind !

Answer 9

So, can you integrate \(3t^2+2t+2\) ?

Answer 10

t^3 + t^2 + 2t

Answer 11

8+4+4
16m/s is the velocity

Answer 12

Almost! You have just forgotten initial velocity is not zero.

Answer 13

:o

Answer 14

why?

Answer 15

Read you qn again.

Answer 16

oh sorry its 2!

Answer 17

v=2+18t
16=2+18t
14=18t?

Answer 18

?????

I said what you did was OK, except you had forgotten that at t=0 , v=2

Answer 19

wait what :/ im confused,where did i go wrong!

Answer 20

You wrote: \(\large v(t) = t^3 + t^2 + 2t\)

you have just forgotten initial velocity (constant of integration):

You should have written: \(\large v(t) = t^3 + t^2 + 2t+V\)

Answer 21

ah! lol sorry :/ so..velocity=18m/s

Answer 22

yep!

Answer 23

18=2+14t
16=14t
lol still getting the same

Answer 24

What are these equations? Haven't you realised your problem is now solved!

Answer 25

thanks!