A motor cycle, starting from rest, has an acceleration of +2.6 m/s^2. When the motorcycle has traveled a distance of 120 m, it slows down with an acceleration of -1.5 m/s^2 until its velocity is +12 m/s. What is the total displacement of the motorcycle?

Answer 1

answer

Answer 2

There are two motions of the journey. The first motion's displacement is given: 120m

Now you calculate the 'Final Velocity' of the first motion to use it as the 'Initial Velocity' of the second motion which we can then use to get the second motion's displacement.
V1 = initial velocity
V2 = final velocity

First motion V2:
V2² = V1² + 2(a)(d)
V2² = (0m/s²) + 2(+2.6m/s²)(120m)
√V2² = √624 m/s
V2 = 24.9800 m/s

Now we find the time for the second motion of the journey.
Let V1 (initial velocity) be the final velocity of the first motion.
(V2 - V1)/a = t
(12 m/s² - 24.9800 m/s²) / -1.5 m/s² = t
t = 8.6533 s

Now we can calculate the second motion's displacement of our journey with the formula:
d = ((V1 + V2)/2) x t
d = ((24.9800 m/s² + 12 m/s²) / 2) x 8.6533 s
d = 160 m

Now add the first and second motion displacements:
160 m + 120 m = 280m

The total displacement of the motorcycle is 280m