jessiepip
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#1
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I understand part a. put ill write it in as I think its useful for part b. which is where I'm stuck. Any help would be great .

A car starting from rest is driven along a horizontal track.
The velocity of the car, vms^-1 ,at time t seconds is modelled by the equation:

v=0.48t^2-0.024t^3 for 0<t<15

a. Find the distance the car travels in the 1st 10 seconds of the journey.
If I'm right I got 120m

b. Find the maximum speed of the car
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pwk123
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Let v(t) = 0.48t^2-0.024t^3 , for 0<t<15
So it's a function for velocity in respect to time therefore lets treat it like a normal function
How do we find maximum or minimum points of a function? [ Have a go then reveal the spoiler]

Spoiler:
Show
Differentiate to get v'(t)

Then at what value will the function mentioned in the spoiled be min or max?


Spoiler:
Show
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Set v'(t) = 0
Solve for t

We will get values for t therefore we plug back into the original v(t) to get a maximum velocity
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blowdriedcrocs
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a) differentiate the equation and place 0 and 10 as the limits
b) Honestly trial and error with the equation, try and plug in any number within the range
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davros
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(Original post by blowdriedcrocs)
a) differentiate the equation and place 0 and 10 as the limits
b) Honestly trial and error with the equation, try and plug in any number within the range
Are you sure that's what you meant to say?
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jessiepip
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(Original post by pwk123)
Let v(t) = 0.48t^2-0.024t^3 , for 0<t<15
So it's a function for velocity in respect to time therefore lets treat it like a normal function
How do we find maximum or minimum points of a function? [ Have a go then reveal the spoiler]

Spoiler:
Show
Differentiate to get v'(t)

Then at what value will the function mentioned in the spoiled be min or max?


Spoiler:
Show
0
Set v'(t) = 0
Solve for t

We will get values for t therefore we plug back into the original v(t) to get a maximum velocity
Thank you that's really helpful
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