calculus question......

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 ...

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.

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. 

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.