Question on Rates of Displacement, Velocity and Acceleration (1 Viewer)

[paddy86]

Q7 (b) (ii) in the 2003 HSC paper.

I'm able to solve (i), but I am confused with (ii). Thanks in advance.

 

[beefywater]

max velocity occurs when acceleration = 0. So u have to diffentiate to find acceleration which u should let =0. Then sub the t value u get back into velocity

 

[gurmies]

Q7 (b) (ii) in the 2003 HSC paper.

I'm able to solve (i), but I am confused with (ii). Thanks in advance.

By inspection, v = 6. This is because the minimum that cos can be is -1. 2-4(-1) = 6, meaning t = pi.

 

[Timothy.Siu]

v=2-cos t
for max velocity..v'=0
v'=-sin t
sin t=0
t=0, pi, 2pi

max...is

t=pi lol
the other ones are minimum.

 

[boxhunter91]

BEst method is to differentiate. Maximum always refers to gradient function=0.
so when dx/dt=0 there is a maximum velocity. When d^2x/dt^2 (dv/dt)=) there is maximum acceleration. Hope this helps good luck.

 

[paddy86]

Maybe I should have clarified. Sorry.

I'm seeking explanation of the method I found in my past papers books (Coroneos).

"Noting that v=2-4cost and that -1≤cost≤1 then at sight, the maximum value of v occurs when cost=-1. This maximum value is 2-4(-1)=6."

 

[Timothy.Siu]

Maybe I should have clarified. Sorry.

I'm seeking explanation of the method I found in my past papers books (Coroneos).

"Noting that v=2-4cost and that -1≤cost≤1 then at sight, the maximum value of v occurs when cost=-1. This maximum value is 2-4(-1)=6."

because the minimum value of cos x is -1.

cosx ranges from -1 to 1
so like the maximum value of -cos t is when cost=-1

 

[untouchablecuz]

so v_max = 6ms^-1

 

[Brontecat]

^_^ I don't understand how the above works

 

[paddy86]

Ok I understand now thanks all.

By sight it refers to the cos graph which has amplitude of 1