Question

A particle moves along a line so that at time t0 its acceleration is given by a(t) = 12t

Answer 1

The value of distance covered by the particle during the first 3 second which moves along a line so that its acceleration is 12t, is 54 ft.

What is the acceleration of an object?

Acceleration of an object is the rate of change of velocity of the object per unit time.

A particle moves along a line so that at time t0 its acceleration is given by $a(t) = 12t\rm\; ft/sec^2$

The value of distance covered by the particle during the first 3 second has to be obtained.

Acceleration is the change in velocity (v) per unit time. Thus, the above equation can be written as,

$a(t) = 12t\rm\; ft/sec^2\\\dfrac{dv}{dt}= 12t\rm\; ft/sec^2$

The equation can be solved as further.

$\dfrac{dv}{dt}= 12t\\v= 6t^2+v_0\rm$

Here, the vo is the initial velocity which is equal to zero.

$v=6t^2$

Velocity is the change in distance(s) per unit time. Thus, the above equation can be written as,

$\dfrac{ds}{dt}=6t^2\\\dfrac{ds}{dt}=2t^3+s_0$

Here, the so is the initial position which is equal to zero. Thus,

$s=2t^3$

Put the value of time t=3.

$s=2(3)^3\\s=54\rm\; ft$

Thus, the value of distance covered by the particle during the first 3 second which moves along a line so that  its acceleration is given by a(t) = 12t is 54 ft.

Learn more about the acceleration here;

https://brainly.com/question/605631

#SPJ4