Question

A particule begins from rest at a point +10 meters from the origin at time t=0, and begins acceleration at a constant 2m/s² in the negative direction, at time t=4seconds the particule has reached a certain speed, it stop traveling and continue with the same speed until t=7seconds. What is its position relative to the origin t=7 seconds

Answer 1

The particle's position relative to the origin at t=7 seconds is -14 meters.

Step-by-step explanation:

Given that the particle starts from rest at a position of +10 meters from the origin, and accelerates at a constant rate of -2m/s² in the negative direction, we can calculate its position at t=7 seconds.

First, we need to find the velocity of the particle at t=4 seconds. Since it starts from rest and accelerates at a rate of -2m/s², we can use the equation v = u + at, where u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we get v = 0 + (-2)(4) = -8 m/s.

Next, we need to find the displacement of the particle from t=4 seconds to t=7 seconds. Since the particle travels at a constant velocity after t=4 seconds, its displacement is equal to the product of its velocity and the time interval. So, the displacement is (-8)(7-4) = -24 meters.

Finally, we add the displacement to the initial position (+10 meters) to find the particle's position relative to the origin at t=7 seconds. Therefore, the particle's position relative to the origin at t=7 seconds is -24 + 10 = -14 meters.