Recent content by bdb1324

Ax = (10-t)m/s^2 Xi = 0 m Vi = 0 m/s t = 0 s since acceleration is A = (Vf-Vi)/t

A particle's acceleration is described by the function a_x =(10 -t) m/s^2, where t is in s. Its initial conditions are x_0 =0 m and v_0x =0 m/s at t =0 s. At what time is the velocity again zero? I am having a hard time setting this problem up with where to begin if someone could help me...

A particle's acceleration is described by the function a_x =(10 -t) m/s^2, where t is in s. Its initial conditions are x_0 =0 m and v_0x =0 m/s at t =0 s. At what time is the velocity again zero? I am having a hard time setting this problem up with where to begin if someone could help...

thanks I got it. It was the same problem. (L*h)/vt^2

isn't v/t the initial starting point of the wall and to get the furthest distance the wall is away wouldn't that be v*t

no since they are similar triangles and v/t is the starting point of h than the larger triangle must be v*t

Homework Statement A small source of light S is located at a distance L from a vertical wall. An opaque object with a height of h moves toward the wall with constant velocity v_vec of magnitude v. At time t= 0 , the object is located at the source S. Find an expression for v_s, the...