The velocity v of a particle moving in the xy plane is given by v = (6.0t - 4.0t
ID: 1499516 • Letter: T
Question
The velocity v of a particle moving in the xy plane is given by v = (6.0t - 4.0t2) i hat + 6.0 j, with v in meters per second and t (> 0) is in seconds.
(a) What is the acceleration when t = 1.0 s? Correct: Your answer is correct. . m/s2 i hat + m/s2 j + m/s2 k
(b) When is the acceleration zero? (Enter 'never' if appropriate.) Correct: Your answer is correct. . s
(c) When is the velocity zero? (Enter 'never' if appropriate.) Correct: Your answer is correct. . s
(d) When does the speed equal 10 m/s? (Enter 'never' if appropriate.) Incorrect: Your answer is incorrect. . s
I need help with part d
Explanation / Answer
given velocity
v = (6.0t - 4.0t2)i +6.0j
(a) acceleration a = Dtv = (6.0 - 8.0t)i +0.0j + 0.0k
at t= 1.0 , a = (6.0 - 8.0)i + 0.0j + 0.0k
= -2.0i m/s2 +0.0j m/s2 + 0.0k m/s2
(b) accleration will be zero
if 6.0 - 8.0t = 0
t= 6.0/8.0 s = 0.75 s
(c) velocity will be zero
if 6.0t - 4.0t2 = 0
t(6.0 - 4.0t) = 0
t= 0.0 and 6.0/4.0 = 0.0 and 1.5 s
(d) speed = (vx2 + vy2 + vz2)^(1/2) = 10
(vx2 + 6.02)^(1/2) = 10
vx = 8.0, - 8.0
if 1) case vx = 8.0
6.0t - 4.0t2 = 8.0
4.0t2 - 6.0t + 8.0 = 0
above equation does not has real solution.
2) case vx =- 8.0
6.0t - 4.0t2 =- 8.0
4.0t2 - 6.0t - 8.0 = 0
t = (-6.0 +(6.02 + 4*4.0*8.0)^(1/2)) / 2*4.0
= 3.4 s
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