IIT JEE , AIEEE Forum - Physics , Mechanics

piyushsahani

If the velocity of the particle is given by u = root of (180 -16x) m/s its acceleration will be __________________ .

gomurali

v =

180 -16 x squaring both sides gives v² = 180 -16x comparing with the equation v² - u² = 2 a x, gives 2 a x = 16x Hence, a = 8 m s^-2 . hope the method is clear. if satisfied pl rate me

krishna.gopal

Good answer.

elessar_iitkgp

v² = 180-16x
Differentiating it twice,
2v(dv/dx) = -16
v(dv/dx) = -8
(dx/dt)(dv/dx) = -8
(dv/dt) = -8
a = -8 m/s²

Gomurali, you have assumed the acceleration to be constant. However, that isn't specified in the problem. So its safer to use the above approach.

And comparison gives, (as gomurali has done)
2a(

x) = -16x
Either

x = 0, or a = -8.
That is the particle's displacement is permanently zero, or its acceleration is -8.
This solution can be useful if the problem mentions that the acceleration is constant. In general, it might help to remember that if v² is a linear function of x, then, the acceleration is constant. (for 1-D motion)

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