Question

A particle has an initial speed of 27 m/s rightward from the origin at t-0. It decelerates at 6t m/s2 (t in seconds). Find the speed after it has traversed 10m of distance. At what time does this occur 1. 2. At t = 0 a particle has a speed of 8 m/s rightward and is located at s-10m. Knowing a 2v3 (v in m/s) find the location and velocity at 4 seconds. What is the maximum arrow speed if its acceleration decreases to zero linearly from A to B. The release value at A is 192,000 in/s2. Determine the time from A to B. 3. Bl 24

Please answer all 3. Thanks

Answer 1

Given that acceleration, a = -6t negative sign indicates because it declaration we know that acceleration, a = is given by, a = dv/dt a - dt = dv integrating both sides we get integral_1^2 dv = integral a. dt v_2 - v_1 = integral a. dt \r\n v_2 - v_1 = integral - 6 t . dt v_2 - v_1 = -6 integral t . dt v_2 - v_1 = -6 t^2/2 given that v_1 = 27 m/sec therefore v_2 - 27 = -6 t^2/2 v_2 - 27 = -3t^2 v_2 = -3t^2 + 27 Rightarrow 1 we know that velocity v = dx/dt \r\n dx = v dt integrating both sides we get integral_1^2 dx = integral v_2 dt [x]_1^2 = integral (- 3 t^2 + 27) dt x_2 - x_1 = integral - 3t^2 dt + integral 27.dt x_2 - x_1 = -3 t^3/3 + 27 t here x_1 = 0 m therefore x_2 = - t^3 +27 t \r\n given that x_2 = 10 m therefore 10 = - t^3 +27 t Rightarrow 2 i.e t^3 - 27 t +10 = 0 on solving we get t_1 = 55 sec, t_2 = 0.372 sec t_3 = -5.372 sec

Answer 2

Given that acceleration, a = -6t negative sign indicates because it declaration we know that acceleration, a = is given by, a = dv/dt a - dt = dv integrating both sides we get integral_1^2 dv = integral a. dt v_2 - v_1 = integral a. dt \r\n v_2 - v_1 = integral - 6 t . dt v_2 - v_1 = -6 integral t . dt v_2 - v_1 = -6 t^2/2 given that v_1 = 27 m/sec therefore v_2 - 27 = -6 t^2/2 v_2 - 27 = -3t^2 v_2 = -3t^2 + 27 Rightarrow 1 we know that velocity v = dx/dt \r\n dx = v dt integrating both sides we get integral_1^2 dx = integral v_2 dt [x]_1^2 = integral (- 3 t^2 + 27) dt x_2 - x_1 = integral - 3t^2 dt + integral 27.dt x_2 - x_1 = -3 t^3/3 + 27 t here x_1 = 0 m therefore x_2 = - t^3 +27 t \r\n given that x_2 = 10 m therefore 10 = - t^3 +27 t Rightarrow 2 i.e t^3 - 27 t +10 = 0 on solving we get t_1 = 55 sec, t_2 = 0.372 sec t_3 = -5.372 sec