an object of mass 3 kg is at rest. Now a force of vectorS = 6t2^i + 4tj^ is applied on the object then velocity of object at t = 3 sec. is?
To find the velocity of the object at t = 3 sec, we need to follow a few steps.
Step 1: Find the acceleration of the object by applying Newton's second law of motion, F = ma. The force acting on the object is given by vector S = 6t^2i + 4tj.
Given that the mass of the object is 3 kg, we have:
F = ma
=> 6t^2i + 4tj = 3a
Step 2: Differentiate the given force equation with respect to time (t) to find the acceleration.
Differentiating 6t^2 with respect to t gives 12t.
Differentiating 4t with respect to t gives 4.
Therefore, our acceleration (a) is:
a = 12ti + 4j.
Step 3: Integrate the acceleration equation with respect to time (t) to get the velocity (v).
Integrating the expression 12ti with respect to t gives us (6t^2)i + C1j, where C1 is the constant of integration.
Integrating 4j with respect to t gives us 4tj + C2i, where C2 is another constant of integration.
Since the object is initially at rest, the initial velocity is zero, so we can substitute v = 0 when t = 0 to find C2.
When t = 0, we have:
v = (6(0)^2)i + C1j + 4(0)j + C2i
0 = 0 + C1j + 0 + C2i
0 = C1j + C2i
Since vectors i and j are perpendicular to each other, their dot product will be zero:
0 = C1j · C2i
0 = C1 * 0 + C2 * 0
0 = 0
Since both C1 and C2 are zero, the equation becomes:
0 = 0j + 0i
0 = 0
So we find that C2 is zero.
Therefore, our velocity equation is:
v = (6t^2)i + 4tj.
Step 4: Substitute t = 3 sec into the velocity equation to find the velocity at t = 3 sec.
When t = 3 sec:
v = (6(3)^2)i + 4(3)j
v = 54i + 12j.
Therefore, the velocity of the object at t = 3 sec is 54i + 12j.
Dun public school
Oops --
v = 2/3 t^3 i + 2/3 t^2 j
FF = ma = 6t^2i + 4tj
Since m=3,
a = 2t^2i + 4/3 t j
so,
v = 2/3 t^2 i + 2/3 t^2 j