Kinematics Question

[isaiguana]

I came across a bit of a strange question on my mock exam today and many people got different answers, so I wanted to see some input on it. I have written the question by memory, so the wording is not entirely exact. Any help is greatly appreciated.

 

The velocity of a particle is v = -1/s², where s is the distance in metres from a fixed point "O". Find the acceleration when the distance from the fixed point is 50 cm.

 

 

 

 

[Ossih]

I'm not sure if this is right but here's my shot

 

v = 1/s^2

dv/dt = d/ds (1/s^2) * ds/dt  [using the chain rule]

a = (-2/s^3)*v

a = (-2/s^3)(1/s^2)

a = -2/s^5

a = -2/(0.5)^2

a = -2/(0.25)

a = -8 m/s^2

 

Hope this helped

[isaiguana]

I'm not sure if this is right but here's my shot

 

v = 1/s^2

dv/dt = d/ds (1/s^2) * ds/dt  [using the chain rule]

a = (-2/s^3)*v

a = (-2/s^3)(1/s^2)

a = -2/s^5

a = -2/(0.5)^2

a = -2/(0.25)

a = -8 m/s^2

 

Hope this helped

 

You've forgotten the negative sign in front of the original equation. 

Nevertheless, attempting your method:

v= -1/s^2

dv/dt= (2/s^3)*ds/dt

dv/dt= (2/s^3)*v

a= (2/s^3)(-1/s^2)

a= -2/s^5

a= -2/(0.5)^5

a= -64

 

Upon graphing the velocity function, I noticed that it becomes less and less negative. Based on this, I believe I can conclude that the acceleration cannot be negative; it must be positive. As such, I have managed to solve this another way in which I do get a positive acceleration, but I am not quite sure if it is correct. Any thoughts would be greatly appreciated.

 

v= -1/s^2 = s/t

So, (-1/s^2)*(1/s) = 1/t

      -1/s^3 = 1/t

 

Now, a = v/t

a = v*(1/t)

a = (-1/s^2)*(-1/s^3)

a = (1/s^5)

a = 1/(0.5)^5

a = 32

[Ossih]

 

I'm not sure if this is right but here's my shot

 

v = 1/s^2

dv/dt = d/ds (1/s^2) * ds/dt  [using the chain rule]

a = (-2/s^3)*v

a = (-2/s^3)(1/s^2)

a = -2/s^5

a = -2/(0.5)^2

a = -2/(0.25)

a = -8 m/s^2

 

Hope this helped

 

You've forgotten the negative sign in front of the original equation. 

Nevertheless, attempting your method:

v= -1/s^2

dv/dt= (2/s^3)*ds/dt

dv/dt= (2/s^3)*v

a= (2/s^3)(-1/s^2)

a= -2/s^5

a= -2/(0.5)^5

a= -64

 

Upon graphing the velocity function, I noticed that it becomes less and less negative. Based on this, I believe I can conclude that the acceleration cannot be negative; it must be positive. As such, I have managed to solve this another way in which I do get a positive acceleration, but I am not quite sure if it is correct. Any thoughts would be greatly appreciated.

 

v= -1/s^2 = s/t

So, (-1/s^2)*(1/s) = 1/t

      -1/s^3 = 1/t

 

Now, a = v/t

a = v*(1/t)

a = (-1/s^2)*(-1/s^3)

a = (1/s^5)

a = 1/(0.5)^5

a = 32

 

 

You can't use v = s/t, because that's average. You need to use v = ds/dt. And say you did graph the velocity function, you'd be graphing it with respect to distance and not time so you can't figure out the acceleration from the graph I think.