# Maths help

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#1
A car is travelling on a straight horizontal road, the velocity of the car v ms^-1 at time t seconds as it passes three point, P Q and R. It is modelled by the equation v=at^2+bt+c
The car passes P at time t=0 with velocity 8ms-1
c=8
The car passes Q at t=5 and at that instant its deceleration is 0.12 ms-2
The car passes R at t=18 with velocity 2.96 ms-1
b) determine the value of a and b
c) find to the nearest metre the distance between points P and R
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#2
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#3
Why do you all hate me ;-;
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#4
*cries*
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#5
0
4 weeks ago
#6
(Original post by hypan6)
Dont worry brother i got you
between q and r it has been 13seconds, 2.96/0.12 = 230
as your initial velocity is 8, a must be 8 and b must be 238
The distance between p and r is 230/13 = 17.7
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4 weeks ago
#7
(Original post by Bruh Moment #293)
Dont worry brother i got you
between q and r it has been 13seconds, 2.96/0.12 = 230
as your initial velocity is 8, a must be 8 and b must be 238
The distance between p and r is 230/13 = 17.7
Yeah, that's what I got
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