Question

An object moves with constant acceleration 3.40 m/s2 and over a time interval reaches a final velocity of 12.6 m/s.

(a) If its original velocity is 6.30 m/s, what is its displacement during the time interval? m

(b) What is the distance it travels during this interval? m

(c) If its initial velocity is −6.30 m/s, what is its displacement during this interval? m

(d) What is the total distance it travels during the interval in part (c)? m

Answer 1

  v=final velocity .. .. u=initial velocity

(a) v² = u² + 2ad
.. d = (v² - u²) / 2a .. .. d = (12.6² - 6.30²) / (2 x 3.40) .. .. ►d = 17.510 m

(b) same as (a) .. as no change in direction is involved in either case.

(c) u = - 6.30m/s.. (assuming final vel. remains at 12.6 m/s)
.. d = (v² - u²) / 2a .. .. d = (12.6² - [-6.30]²) / 2a .. .. d = 119.07 / 6.8 .. .. ►d = 17.51 m

(d) Distance to decelerate -6.30 m/s to 0 m/s = d1
... Distance to accel. 0 m/s to 12.6 m/s = d2

d1 = (v² - u²) / 2a .. d1 = (0² - [-6.30]²) / (2 x 3.40) .. .. d1 = (-) 5.836 m .. ('backwards')

d2 = (v² - u²) / 2a .. d2 = (12.6² - 0²) / (2 x 3.40) .. d2 = 23.347 m

Total distance = d1 + d2 .. .. D = 5.836 + 23.347 .. .. ►D = 29.183 m