A particle moves along the path y=x^3+3x+1 where units are in centimetres.If the horizontal velocity Vx is constant at 2cm/s,find the magnitude and direction of the velocity of the particle at the point (1,5).
To find the magnitude and direction of the velocity of the particle at the point (1,5), we need to evaluate the derivative of the position function with respect to time.
Given that the particle's position is given by y = x^3 + 3x + 1, we can obtain its velocity by taking the derivative of y with respect to time.
Let me explain how to find the derivative of y with respect to time:
1. Express y in terms of t instead of x:
y = x^3 + 3x + 1
2. Since the horizontal velocity Vx is constant at 2cm/s, we can express x in terms of t:
x = 2t
3. Substitute x = 2t back into the position function:
y = (2t)^3 + 3(2t) + 1
= 8t^3 + 6t + 1
4. Differentiate y with respect to t to find the velocity function:
dy/dt = d/dt (8t^3 + 6t + 1)
Differentiating, we get:
dy/dt = 24t^2 + 6
Now, we have the velocity function dy/dt = 24t^2 + 6.
To find the magnitude and direction of the velocity at the point (1,5), we need to evaluate the velocity function at t = 1.
Let me find the velocity at t = 1:
1. Substitute t = 1 into the velocity function:
dy/dt = 24(1)^2 + 6
= 24 + 6
= 30 cm/s
So, the magnitude of the velocity at the point (1,5) is 30 cm/s.
Next, to find the direction of the velocity at this point, we need to determine whether the particle is moving in the positive or negative y-direction.
Let me determine the direction of the velocity at (1,5):
1. Evaluate the sign of dy/dt at t = 1:
dy/dt = 24(1)^2 + 6
= 30 cm/s
Since dy/dt is positive at t = 1 (30 cm/s is positive), the particle is moving in the positive y-direction.
Therefore, the magnitude of the velocity at the point (1,5) is 30 cm/s, and the direction is in the positive y-direction.
dy/dt = 3x^2+3 dx/dt
so, since Vx=dx/dt=2,
dy/dt = 6(x^2+1)
at x=1, Vy=dy/dt = 12
so, V=√(4+144)
tanθ = Vy/Vx = 6