Question

1. (12 points) this is a multiple part question. draw a particle diagram for each section of travel described below. for parts a. - c. clearly indicate the known values and what you are looking for. show the equations without numbers as your first step. box your final answers.

a car starts from rest at a stop sign. it accelerates at 2.0 m/s² for 6.0 s. the driver takes her foot off the accelerator and then coasts (maintains the same speed) for 2.0 s.
draw the particle diagram on the line below. ↓ use the line as an x axis.
a. at the end of the acceleration period, how fast is the car going?
b. now determine how far the car travels during its acceleration period.
c. finally, how far does the car travel during the coasting period?

Explanation:

Step1: Identify the initial - velocity and acceleration formula

The car starts from rest, so $v_0 = 0\ m/s$. The formula for velocity in uniformly - accelerated motion is $v=v_0 + at$.

Step2: Calculate the velocity at the end of the acceleration period

Substitute $v_0 = 0\ m/s$, $a = 2.0\ m/s^2$, and $t = 6.0\ s$ into the formula $v=v_0+at$.
$v=0 + 2.0\times6.0=12\ m/s$

Step3: Identify the formula for displacement in uniformly - accelerated motion

The formula for displacement in uniformly - accelerated motion is $x=v_0t+\frac{1}{2}at^{2}$.

Step4: Calculate the displacement during the acceleration period

Substitute $v_0 = 0\ m/s$, $a = 2.0\ m/s^2$, and $t = 6.0\ s$ into the formula $x=v_0t+\frac{1}{2}at^{2}$.
$x=0\times6.0+\frac{1}{2}\times2.0\times6.0^{2}=36\ m$

Step5: Identify the formula for displacement in uniform - motion

In the coasting (uniform - motion) period, the speed $v = 12\ m/s$ (from part a) and the formula for displacement is $x = vt$.

Step6: Calculate the displacement during the coasting period

Substitute $v = 12\ m/s$ and $t = 2.0\ s$ into the formula $x = vt$.
$x=12\times2.0 = 24\ m$

Answer:

a. $12\ m/s$
b. $36\ m$
c. $24\ m$