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natima [27]
3 years ago
8

An object moves such that acceleration a(t)=2(1-3t). Given v(0)=-5 and x(0)=4, find x(1).

Mathematics
1 answer:
Vesnalui [34]3 years ago
5 0

Answer:

x(1) = -1

Step-by-step explanation:

Given the equation for acceleration of the object as;

a(t)=2(1-3t)

a(t) = 2-6t

Integrate to get the velocity

v(t) = 2t-6t²/2 + C

v(t) = 2t-3t²+C

If v(0) = -5

when t = 0 v = -5

-5 = 2(0)-3(0)²+C

-5 = C

C = -5

v(t) = 2t-3t²-5

To get the displacement x(t), we will integrate the velocity function

x(t) = 2t²/2-3t³/3-5t+C

x(t) = t²-t³-5t+C

If x(0) = 4

At t = 0, x(t) = 4

4 = 0²-0³-5(0)+C

4 = C

C = 4

x(t) = t²-t³-5t+4

Get x(1)

x(1) = 1²-1³-5(1)+4

x(1) = 1-1-5+4

x(1) = -5+4

x(1) = -1

Hence x(1) = -1

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y-(-1)=\dfrac{3}{5}(x-(-3))\\\\y+1=\dfrac{3}{5}(x+3)\qquad\text{use distributive property}\\\\y+1=\dfrac{3}{5}x+\dfrac{9}{5}\qqua\text{subtract 1 from both sides}\\\\y=\dfrac{3}{5}x+\dfrac{4}{5}\qquad\text{multiply both sides by 5}\\\\5y=3x+4\qquad\text{subtract 3x from both sides}\\\\-3x+5y=4\qquad\text{change the signs}\\\\3x-5y=-4

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