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calculus question......

the position function s of a point is given by two dimensional quantities  where t in seconds and the y direction is affected by gravity (10m/s/s).

y(t)=4t*sin(θ) - 5t2       &       x(t)=4t*cos(θ)

1- if θ=pi/3, find the speed in each direction after 1 second.

2- find the acceleration after 2 seconds. does it vary with time?

Please any help is welcome....

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Brad M. | STEM Specialist plus Business, Accounting, Investment & EditingSTEM Specialist plus Business, Accountin...
4.9 4.9 (229 lesson ratings) (229)
1

Hi Ted -- Good to hear from someone in Virginia!

This is a projectile problem that you can sense physically:

Initial velocity launched up at 4 m/s and / 60 degrees, with horizontal component constant x'(0)= 2 m/s and

vertical part y'(0)= 3.7 m/s fighting a constant negative acceleration 10 m/s^2.

After one second, y'(1) suffers a 10 m/s reversal from the starting upward 3.7 ==> -6.3 m/s (downward). Horizontal x'(1) speed remains constant 2 m/s, unchanged by gravity or air drag (assumed).

After two seconds, vertical y"(2) is always -10 m/s/s -- gravity's constant "pull". There can be no horizontal speed change ==> no acceleration ever, x"=0.  Warmest regards, sir ...

Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
1

1. y'(1) = 4sin(pi/3) - 10 = 2sqrt(3) - 10; x'(1) = 4cos(pi/3) = 2

 

2. y''(2) = -10; x''(2) = 0. So, the acceleration doesn't very with time.

Martin S. | Mathematics and Physics Tutor For HireMathematics and Physics Tutor For Hire
0

Greetings Ted. 

Speed is a scalar quantity. So it's essentially positive. 

cos (pi/3) = 1/2; sin (pi/3) = sqrt(3)/2

Velocity (x-direction) = vx = dx/dt = d/dt (4*t*cos(pi/3) = 4*(1/2) = 2 m/s

Since the velocity is positive, the velocity is also the speed.

Acceleration (x-direction)  = ax = d(vx)/dt = 0

The horizontal velocity is constant.

Velocity (y-direction) = vy = dy/dt = d/dt (4*sin(pi/3)*t - 5*t^2) = 4*(sqrt(3)/2) - 10*t

After 1 second, vy = 2*sqrt(3) - 10(1) = -6.54 m/s

Since we're asking for speed (velocity is a vector), we must take the absolute value

Vertical Speed = abs(vy) = 6.54 m/s (downward)

Acceleration (y-direction) = ay = d(vy)/dt = -10 m/s^2

Neglecting air resistance, the gravitational acceleration is constant. 

Ergo, acceleration does not with time. 

Hope this helps. 

Grigori S. | Certified Physics and Math Teacher G.S.Certified Physics and Math Teacher G.S.
0

There is no change in the horizontal direction since the force of gravity is acting straight down and doesn't have its x - component. That is why the work done by gravity along x axis is zero.