How to solve Position and Time questions...?

Question

If a particle's position is given by:

x=4m-(12m/s)t+(3m/s^2)t^2 (where t is in seconds and x is in meters), what is its velocity at t1 = 1s?

Answer 1

Differentiate the position function with respect to time & you will get the velocity.

Here, velocity, v = dx/dt

                            = -12m/s + (6m/s^2)t

At t1 =1s

v = -12m/s + (6m/s^2) X 1s

  = -6m/s

Answer 2

a particle's position is given by:

x=4m-(12m/s)t+(3m/s^2)t^2

=

x=4m-vt+at^2

=

x=4-12t+3t^2

velocity= dx/dt

0-12+6t

when t=1

v=-6 m/s

Answer 3

GIVEN

x=4m-(12m/s)t+(3m/s^2)t^2

the velocity is the derivative of the above, i.e.,

dx/dt = - 12 + 6t

and at t = 1,

dx/dt = -12 + 6(1)

dx/dt = -6 m/sec

Hope this helps.

Answer 4

right now u said x is the position, differentiate with respect to t and ull have speed differentiate again and ull get acceleration