A particle P moves in a straight line such that at time T seconds its displacement, S meters from a fixed point O on the line is given by
S=T^3-7T^2+10T T>0
a) find the values of t (t>0) at which P passes through O.
b)find the speed of P each time it passes through O.
c) Find the greatest speed of P in the interval 0<T<5
Brian
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a) S = T^3 - 7*T^2 + 10*T = T*(T^2 - 7*T + 10) = T*(T - 2)*(T - 5) = 0 when
T = 0, T = 2 and T = 5 seconds.
b) dS/dT = 3*T^2 - 14*T + 10, which at
(i) T = 0 seconds is 10 units/s,
Iii) T = 2 seconds is -6 units/s,
(iii) T = 5 seconds is 15 units/s.
c) d/dT(dS/dT) = 6*T - 14 = 0 when T = 14/6 seconds, at which time
dS/dT = 3*(14/6)^2 - 14*(14/6) + 10 =
(588 - 1176 + 360)/36 = -228/36 = -6 1/3 units/s, which is a minimum
on the interval. We also need to look at the endpoints, which we already
checked in b), where we found that dS/dT at T = 5 was 15 units/s, and so this
would be the greatest speed of P on (0,5).
Polygon
S=T^3-7T^2+10T T>0
a) find the values of t (t>0) at which P passes through O.
When P passes through O S = 0
0=T(T - 2)(T - 5)
P passes through O at 0, 2 and 5 seconds
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b) find the speed of P each time it passes through O.
S=T^3-7T^2+10T
Speed = dS/dT = 3T^2 - 14T + 10
When T = 0, speed = 10 m/sec
When T = 2, speed = -6 m/sec
When T = 5, speed = + 5 m/sec
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c) Find the greatest speed of P in the interval 0<T<5
At Max or Min dS/dT= 0
procter
nicely, dy/dx is the gradient of a line, so the gradient of that line at any factor may well be dy/dx of that function. to discover the by-fabricated from a function you basically use a formula nx^n-a million = dy/dx of x^n, so which you may desire to be conscious this 3 . 2/3 . x^2 + 2 . x - 4 Then : 2x^2 + 2x - 4 = dy/dx, At table sure factors the gradient might desire to equivalent 0, because of the fact there is not any gradient on the table sure factor, think of approximately it, at a flat piece of land is the land increasing or reducing, the respond is its no longer, so dy/dx at table sure factors. so 0 = 2x^2 + 2x - 4, and you recognize a thank you to factorize, (nicely you may desire to) 2(x^2 + x -2) (extra 2 outdoors) 2((x+2)(x-a million)) = 0 So the only suggestions for this are x = -2 and x = a million, to discover the character of those table sure factors use the 2d derivative attempt, so which you come across the 2d by-fabricated from the function, basically like the style you discovered the by-fabricated from the 1st function, So: 2x + 2 = d^2y/dx^2 And whilst the 2d derivative is unfavourable the table sure factor is a optimal and vice versa. so which you sub the numbers in; 2(-2) + 2 = -2, subsequently that's a max, and a million(2) + 2 = 4, subsequently that's a min. Now you basically sub the numbers you won to get the y co-ordinate, 2/3 (-2)^3 + (-2)^2 - 4(-2) = 6/3 and a pair of/3 + a million - 4 = -7/3 There you pass!