Question

HELPPPPP

Answer 1

@Destinymasha

Answer 2

@NERD85176

Answer 3

Do you know your kinematic equations of motion for constant acceleration?

Does \(x = x_0 + v_0t + {1 \over 2 } at^2\) ring a bell?

Answer 4

Velocity is defined as displacement per unit time.

For example, 60 miles per hour is a typical highway velocity for a car. This means that we cover 60 miles in one hour of travel.

\[v = {\Delta d \over \Delta t} \left [{\rm distance \over time} \right ]\]

Acceleration is defined as the change in velocity per unit time.

For example, if we accelerate from 0 to 10 meters per second in 1 second, our acceleration would be \(10 m/s^2\)\[a = {\Delta v \over \Delta t} \left [ \rm velocity \over time \right ] \left [ \rm m/s \over s \right ] \left [\rm m \over s^2 \right ]\]

If you haven't done calculus, you'll just have to take my word for the following.

Commit these equations to memory, as you'll use them plenty in your physics class. These are called the kinematic equations of motion for constant acceleration. This means acceleration doesn't change with time. For example, if a car stops from 60 miles per hour to 0 miles per hour with a deceleration of 5 milers per hour per second, the car will have constant acceleration (or in this case, constant deceleration).

The distance travelled by an object at initial position \(x_0\), with initial velocity \(v_0\), and constant acceleration \(a\) over a time interval \(t\) will be \[x = x_0 + v_0t + {1 \over 2} a t^2\]

Here are the rest of the equations: \[v = at + v_0\]\[v^2 = v_0^2 + 2 a (r-r_0)\]

We use each of these three depending on which values we are given in our problem statement.

In this case, we ignore the first part of the problem. We are just solving for the distance travelled by an object whose initial velocity is 30 m/s and is decelerating at a rate of \(0.065 m/s^2\).

We want the use the following equation: \[v^2 = v_0^2 + 2 a (x-x_0)\]

The initial velocity, \(v_0\) is 30.
The final velocity, \(v\) is 0.
The acceleration, \(a\) is -0.065. (It is negative because we are slowing down or decelerating.)
The initial position, \(x_0\) is 0.

We can now solve for the distance travelled by the slowing train, \(x\).