HURRY PLEASE SHOW THE WORK TOO!! A truck and a motorbike are at rest and are 900 m apart. They start moving toward, the car accelerates at a rate of 3.2 m/s2 and the motorcycle accelerates at a rate of 5.1 m/s2. At what position will each object pass each other relative to the starting point?

They pass each other at 2415.79 m relative to the starting point.

Explanation:

Let s represent the distance moved by the car.

Since they are 900 m apart, it starts from rest and the truck moves at an acceleration of a = 3.2 m/s², we use

s - s₀ = ut + 1/2at²      u = initial velocity = 0 (since it starts from rest) and s₀ = 900

s = 900 + 1/2 × 3.2 m/s² × t²

s = 900 + 1.6t²

Let s' represent the distance moved by the motorcycle.

Since it starts from rest and the motorcycle moves at an acceleration of a = 5.1 m/s², we use

s' - s₀' = ut + 1/2at²      u = initial velocity = 0 (since it starts from rest) and s₀' = 0 (the starting point of the motorcycle)

s' = 0 + 1/2 × 5.1 m/s² × t²

s' = 2.55t²

At the point where they pass each other, s' = s

2.55t² = 900 + 1.6t²

collecting like terms, we find the time which they meet each other

2.55t² - 1.6t² = 900

0.95t² = 900

t² = 900/0.95

t = √(947.37)

t = 30.78 s

Substituting t into s', we have

s' = 2.55t²

= 2.55(30.78)²

= 2.55(947.37)

= 2415.79 m

So, they pass each other at 2415.79 m relative to the starting point.

gravity

explanation:

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