Question

An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 seconds. Then, a constant acceleration of 7 m/s2 is applied to it in the +x direction for 9 seconds. What is the total distance covered by this object in meters? Please give a detailed explanation.

Answer 1

244.64m

Step-by-step explanation:

First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:

`x = V*t = -8(m)/(s) *3s = -24m`

After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

`v^2 = v_0^2 + 2a*d`

Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

`d = (v^2 - v_0^2)/(2a) = ((0m/s)^2 - (-8m/s)^2)/(2*7m/s^2)= -4.57m`

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

`t_1 = (v)/(a) = (8m/s)/(7m/s^2) = 1.143 s`

Then, the time of the second phase will be:

`t_2 = 9s - t_1 = 9s - 1.143s = 7.857s`

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

`x = (1)/(2)a*t^2 + v_0*t + x_0`

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

`x = (1)/(2)*7(m)/(s^2)*t^2 + 0m/s*t + 0m = 216.07m`

So, the total distance covered by this object in meters will be the sum of all the distances we found:

`x_total = 24m + 4.57m + 216.07m = 244.64m`