Question

3 t If x=3 t^3 - 7t²+ 6t-11 is the displacement then calculate velocity at 1 second & 2 seconds ?​

Answer 1

Answer:

The velocity at time 1 second is -9 units per second and the velocity at time 2 seconds is -1.5 units per second.

Explanation:

The function of displacement 'x' in terms of time 'y' is represented as-

x= 3 t^3 - 7t^2 + 6t - 11

So displacement at time t = 1 second is;

x(1) = 3 x 1^3 - 7 x 1^2 + 6x1 - 11

= -9

Displacement at time t = 2 seconds is;

x(2) = 3 x 2^3 - 7 x 2^2 + 6x2 - 11

= -3

So, velocity at time 1 second is;

= displacement ÷ time

= (-9) ÷ 1

= -9 units per second

Velocity at time 2 seconds is;

= displacement ÷ time

= (-3) ÷ 2

= -1.5 units per second

( minus sign depicts the negative quantity as velocity is a vector quantity)