An electron's position is given by r = 4.00t i - 6.00t2 j + 5.00 k, with t in seconds and r in meters.
(a) In unit-vector notation, what is the electron's velocity v(t)? (Answer in terms of i, j, k, and t.)
___ m/s
(b) What is v in unit-vector notation at t = 8.00 s?
___ m/s
(c) What is the magnitude of v at t = 8.00 s?
___ m/s
(d) What angle does v make with the positive direction of the x axis at t = 8.00 s?
___ ° (from the +x axis)
..for (a), i got:
((d(4.00))/dt)i-((d(6.00))/dt^2)j+((d(5.00))/dt)k
and for (b):
((d(4.00))/(d(8.00)))i-((d(6.00))/(d(8.00^2)))j+((d(5.00))/(d(8.00)))k
what am i doing wrong?
To find the velocity of the electron, you need to differentiate the position vector with respect to time (t).
(a) To find the velocity v(t) in unit-vector notation, differentiate each component of the position vector r separately with respect to t:
v(t) = (d/dt)(4.00t)i - (d/dt)(6.00t^2)j + (d/dt)(5.00)k
(d/dt) represents the derivative with respect to time.
Differentiating, we get:
v(t) = 4.00i - 12.00tj + 0k
So, the velocity vector is v(t) = 4.00i - 12.00tj.
(b) To find v at t = 8.00s, substitute t = 8.00s into the velocity vector:
v(8.00s) = 4.00i - 12.00(8.00)j + 0k
v(8.00s) = 4.00i - 96.00j
So, the velocity vector at t = 8.00s is v(8.00s) = 4.00i - 96.00j.
(c) The magnitude of the velocity vector is given by:
|v(8.00s)| = sqrt((v_x)^2 + (v_y)^2 + (v_z)^2)
Substituting the values of v at t = 8.00s, we get:
|v(8.00s)| = sqrt((4.00)^2 + (-96.00)^2 + 0^2)
|v(8.00s)| = sqrt(16.00 + 9216.00)
|v(8.00s)| = sqrt(9232.00)
|v(8.00s)| = 96.06 m/s (approximately)
So, the magnitude of v at t = 8.00s is approximately 96.06 m/s.
(d) To find the angle v makes with the positive direction of the x-axis, we can use trigonometry. The angle can be found using the equation:
tan(theta) = (v_y/v_x)
Substituting the values of v at t = 8.00s, we get:
tan(theta) = (-96.00)/(4.00)
theta = atan(-96.00/4.00)
Calculating the arctan, we get:
theta ≈ -89.4° (approximately)
So, the angle v makes with the positive direction of the x axis at t = 8.00s is approximately -89.4° (from the +x axis).