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.
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 [tex]a(t) = 12t\rm\; ft/sec^2[/tex]
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,
[tex]a(t) = 12t\rm\; ft/sec^2\\\dfrac{dv}{dt}= 12t\rm\; ft/sec^2[/tex]
The equation can be solved as further.
[tex]\dfrac{dv}{dt}= 12t\\v= 6t^2+v_0\rm[/tex]
Here, the vo is the initial velocity which is equal to zero.
[tex]v=6t^2[/tex]
Velocity is the change in distance(s) per unit time. Thus, the above equation can be written as,
[tex]\dfrac{ds}{dt}=6t^2\\\dfrac{ds}{dt}=2t^3+s_0[/tex]
Here, the so is the initial position which is equal to zero. Thus,
[tex]s=2t^3[/tex]
Put the value of time t=3.
[tex]s=2(3)^3\\s=54\rm\; ft[/tex]
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;
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