ok, suppose a car travels around a circular track of radius 100 m. suppose the speed of the car varies in time as v=2 (m/s^2) t. Find the time t at which the angle between the velocity vector and the acceleration vector is 45 degrees.
I have been trying to solve this question for a while and I can not seem to figure it out.
Any hints solution, anything please let me know!
ASAP :D thanks
I have been trying to solve this question for a while and I can not seem to figure it out.
Any hints solution, anything please let me know!
ASAP :D thanks
-
The velocity vector is tangent to the circular track (along the track). The acceleration vector is the sum of two acceleration components: the tangential acceleration and the centripetal acceleration.
The tangential acceleration is d/dt Vt(t) = d/dt (2*t) = 2 m/s²; The centripetal acceleration (at right angles to the tangential) is v²/R = 4*t²/R. The vector sum is the total acceleration and makes an angle of arctan[(4*t²/R)/2] = arctan(2*t²/R) with the tangential. Since the velocity vector is along the tangential, this is also the angle with the velocity vector.
arctan(2*t²/R) = 45º
2*t²/R = 1.0
t² = R/2
t=-√50 = 7.07 s
The tangential acceleration is d/dt Vt(t) = d/dt (2*t) = 2 m/s²; The centripetal acceleration (at right angles to the tangential) is v²/R = 4*t²/R. The vector sum is the total acceleration and makes an angle of arctan[(4*t²/R)/2] = arctan(2*t²/R) with the tangential. Since the velocity vector is along the tangential, this is also the angle with the velocity vector.
arctan(2*t²/R) = 45º
2*t²/R = 1.0
t² = R/2
t=-√50 = 7.07 s
-
try using polar co-ordinates and dot product of vectors to solve this...it'll turn out pretty simple :)