Question

For the following described motion, fill in the position/time, a velocity/time, and acceleartion/time graphs on the grids provided. Show your work for calculating exact positions, velocities and times in the space below. The graphs should be excact.

1. The particle begins at rest at the origin and undergoes an acceleartion of 1m/s for 1s.

2. The acceleration of the particle then changes instantly to zero for 2s.

3. The parrticle then instantly undergoes an acceleration of minus 2m/s for 1s.

4. At this point, the particle comes to rest steadily over a period of 2s.

a. Draw the position/time graph of the motion.

b. Draw the velocity/time graph of the motion.

c. Draw the acceleartion/time graph for the motion.

Answer 1

1.) Position at end of 1 s = 0.5 * a * t^2 = 0.5 * 1 * t^2 = 0.5 *t^2
After 1 s , position = 0.5 * 1^2 = 0.5 m
Velocity = a* t = 1*t = t
After 1 s, velocity = 1*1 = 1 m/s
Acceleration during 1 s = 1 m/s^2
2)
acceleration from t = 1 s to t= 3s = 0
Velocity during this time = 1 m/s
Position during this time is given by = 0.5 + (1 * (t-1)) =t - 0.5
Position at t=3s = 3 -0.5 = 2.5 m
3)
Now acceleration for t= 3 to t= 4s is -2 m/s^2
Velocity during this time = 1+ (-2*(t-3)) = (-2*t) + 7
Velocity at t=4 = -1 m/s
Position during this time = 2.5 + (1*(t-3)) - (0.5 * 2* (t-3)^2) = -t^2 + 7t -9.5
Position at the end of t = 4s = 2.5 m
4) For the particle to come to rest in 2 s
acceleration = V/t = 1/2 = 0.5 m/s^2
Acceleration from t = 4 to t =6 s is 0.5 m/s^2
Velocity during this time = -1 + (0.5 * (t-4)) = (0.5 *t) -3
Velocity at t= 6s = 0 m/s
Position during this time= 2.5 + (-1 *(t-4)) + (0.5 * 0.5 * (t-4)^2) = 0.25t^2 -3t + 10.5
Position at end t= 6s =1.5 m