A particle p moves in a straight line with a velocity v m/sec that is given by v=4t^2-1 where t is the time after passing through a fixed point O on the line .find the instant when p is instantaneously at rest .also find at that instant ,displacement of p from O.
to be at rest, v = 0
4t^2 - 1 = 0
4t^2 = 1
t^2 = 1/4
t = 1/2
let the displacement be s
s = (4/3)t^3 - t + c
assume that when t = 0 , s=0 , then c =0
s= (4/3)t^2 - t
when t = 1/2
s = (4/3)(1/4) - 1/2
= 1/3 - 1/2
= -1/6
the object is 1/6 m to the "left" of O
To find the instant when particle P is instantaneously at rest, we need to find the time when the velocity v is equal to zero.
Given Equation: v = 4t^2 - 1
Setting v to zero, we get:
0 = 4t^2 - 1
Adding 1 to both sides, we have:
1 = 4t^2
Dividing both sides by 4, we get:
1/4 = t^2
To solve for t, we'll take the square root of both sides:
√(1/4) = √t^2
1/2 = t
So, the particle P is instantaneously at rest at t = 1/2 seconds.
Now, to find the displacement of P from point O at that instant, we need to find the position function using the given velocity function.
The position function (s) is the integral of the velocity function (v) with respect to time (t).
So, let's integrate the velocity function:
v = 4t^2 - 1
Integrating with respect to t, we get:
s = ∫ (4t^2 - 1) dt
Integrating term by term:
s = (4/3)t^3 - t + C
Where C represents the constant of integration.
To find the value of C, we can use the additional information that particle P passes through point O.
Since point O is the starting point, the displacement (s) at t = 0 should be zero.
Plugging t = 0 into the position function:
0 = (4/3)(0)^3 - 0 + C
0 = 0 + C
C = 0
Now, substituting t = 1/2 and C = 0 into the position function:
s = (4/3)(1/2)^3 - (1/2) + 0
s = (4/3)(1/8) - (1/2)
s = 1/6 - 1/2
s = -1/3
Therefore, at the instant when P is instantaneously at rest (t = 1/2 seconds), the displacement of P from point O is -1/3 units.