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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. | Summer Online Finance Specialist: WACC NPV DCF TVM YTMSummer Online Finance Specialist: WACC ...
4.9 4.9 (233 lesson ratings) (233)

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. 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

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.

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.