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3 years ago

The displacement (in meters) of a particle moving in a straight line is given by the equation of motion

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

4 0

Answer:

-2/a³ m/s
-2 m/s
-1/4 m/s
-2/27 m/s

Step-by-step explanation:

The velocity is the derivative of position:

v = ds/dt = (d/dt)(t^-2) = -2t^-3

For t=a, the velocity is

-2a^-3 = -2/a³ . . . . meters per second

For t=1, the velocity is ...

-2·1³ = -2 . . . . meters per second

For t=2, the velocity is ...

-2·2^-3 = -2/8 = -1/4 . . . . meters per second

For t=3, the velocity is ...

-2·3^-3 = -2/27 . . . . meters per second

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Answer:

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Step-by-step explanation:The formula concludes to a quadratic function.

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3 years ago

Arnold ran 50 feet. Which equation can be used to determine c Arnold’s total distance in cm if 1 inch is equal to 2.54 centimete

bagirrra123 [75]

Answer:

1,524 centimeters

Step-by-step explanation:

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3 years ago

Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl

GaryK [48]

Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane $n=(-2,-2,1)$ , then r will have the next parametric equations:

$x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda$

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

$-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3$

Substitute the value of $\lambda$ in the parametric equations:

$x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\$

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

$d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u$

7 0

3 years ago

I need help on this question!

lord [1]

Answer:

If A = 8, B = 5

If A = 20, B = 3

Step-by-step explanation:

4x²(2x³ + 5x)

8x⁵ + 20x³

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If A = 20, B = 3

4 0

3 years ago

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Darya [45]

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3 years ago