Question

Two particles A and B start from rest at the origin x = 0 [ft] and move along a straight line such that a = (613) Ift/s) and ag = (1212.8) [ft/s), where t is in seconds. Determine the distance between them when t3 [s), and the total distance each has traveled in 3 seconds

Answer 1

Answer:

Distance between them after 3 s is 2695.5 ft.

Total distance traveled by A in 3 s is 2758.5 ft.

Total distance traveled by B in 3 s is 5454 ft.

Explanation:

For particle A:

u = 0, a = 613 ft/s

Let the distance traveled by particle A in 3 seconds is Sa.

Use second equation of motion

S = u t + 1/2  at ^2

Sa = 0 + 1/2 x 613 x 3 x 3 = 2758.5 ft

For particle B:

u = 0, a = 1212.8 ft/s

Let the distance traveled by particle B in 3 seconds is Sb.

Use second equation of motion

S = u t + 1/2  at ^2

Sb = 0 + 1/2 x 1212 x 3 x 3 = 5454 ft

Thus, the difference in the distance traveled by A and B is 5454 - 2758.5 = 2695.5 ft.