Particle moving under influence of a constant force is given by V =√|4-2X| where X is magnitude of displacement of the particle At t=0 initially the particle is noticed to be moving towards east.The distance travelled by the particle in first 5seconds is?
The ratio of magnitude of displacement in 4th and 5th second is?
so if sqrt(abs(4-2x) is displacement, you will manually change signs at x=2 of the integrand when integrating from fourth to fifth second.
a) distance= int(v dt) 0,2 + (-int(vdt)2,5
and do similar for t=4, and t=5, and remember the signs when you subtract distance@4 from distance@5
To find the distance traveled by the particle in the first 5 seconds, we need to integrate the velocity function over the specified time interval.
Given that the velocity of the particle is V = √|4 - 2X|, we'll integrate this equation with respect to time (t) from 0 to 5.
Let's break down the problem into two intervals - when the particle is moving towards the east (X > 0) and when it is moving towards the west (X < 0).
For X > 0:
Since the particle is moving towards the east initially, the magnitude of displacement (X) is given as X = Vt. Replacing the given velocity function, we have X = √|4 - 2X| * t.
To find the exact value of X, we'll square both sides of the equation:
X^2 = (4 - 2X) * t^2
Rearranging the equation:
2X^2 + 2Xt^2 - 4t^2 = 0
Solving this quadratic equation, we find that the magnitude of displacement X = t * (√(1 + 2t^2) - 1).
Now, let's consider the interval when X < 0:
Since the magnitude of displacement cannot be negative, we'll integrate the equation V = -√|4 - 2X| with respect to t from 0 to 5. This will give us the magnitude of displacement for this interval.
Integrating, we have X = -t * (√(1 + 2t^2) - 1).
To find the total distance traveled, we add the magnitude of displacement for both intervals: X = t * (√(1 + 2t^2) - 1) - t * (√(1 + 2t^2) - 1).
Substituting t = 5, we can calculate the distance traveled by the particle in the first 5 seconds.
For the ratio of magnitude of displacement in the 4th and 5th second, we need to find the displacement at both time points.
Substitute t = 4 in the equation X = t * (√(1 + 2t^2) - 1) to find the magnitude of displacement in the 4th second.
Substitute t = 5 in the same equation to find the magnitude of displacement in the 5th second.
Finally, divide the magnitude of displacement in the 4th second by the magnitude of displacement in the 5th second to find the desired ratio.