Click to Chat
1800-2000-838
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
The position of a particle moving along the x axis depends on the time according to equation x = 3t2-2t3 At what time does the particle reach its maximum positive x position? And when does it reach minimum? (I believe this requires calculus. Could someone show me?)
The position of a particle moving along the x axis depends on the time according to equation x = 3t2-2t3
At what time does the particle reach its maximum positive x position? And when does it reach minimum?
(I believe this requires calculus. Could someone show me?)
Hi Pritish, Remember that at the point of maxima and minima, dy/dx = 0 and if d2y/dx2 is -ve, then that value of x gives maxima and if, d2y/dx2 = +ve, then it is minima. Given, x = 3t2 - 2t3 dx/dt = 6t - 6t2 For maxima, or minima, dx/dt = 0 6t - 6t2 = 0 t = t2 t (t - 1) = 0 t = 0s & t = 1 sec Let us find d2x/dt2 (d2x/dt2 )t=0 = 6 - 12t = 6 - 12 x 0 = 6+ve (d2x/dt2 )t=1 = 6 - 12 x 1 = -6 = - ve Hence, at t = 1 sec, particle reaches its maximum +ve x position and at x = 0 it reaches minimum x - position
Hi Pritish,
Remember that at the point of maxima and minima, dy/dx = 0
and if d2y/dx2 is -ve, then that value of x gives maxima and if,
d2y/dx2 = +ve, then it is minima.
Given, x = 3t2 - 2t3
dx/dt = 6t - 6t2
For maxima, or minima,
dx/dt = 0
6t - 6t2 = 0
t = t2
t (t - 1) = 0
t = 0s & t = 1 sec
Let us find d2x/dt2
(d2x/dt2 )t=0 = 6 - 12t
= 6 - 12 x 0
= 6+ve
(d2x/dt2 )t=1 = 6 - 12 x 1
= -6
= - ve
Hence, at t = 1 sec, particle reaches its maximum +ve x position and at x = 0 it reaches minimum x - position
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -