2. A car accelerates uniformly from +10.0 m/s to +40.0 m/s over a distance of 125 m. How long did it take to go that distance? Show all your work, including the equation used, given and unknown quantities, and any algebra required. Make sure your answer has the correct number of significant figures.

G:vi=10.0, vf=40.0, deltax=125

U:time
E:deltax=1/2(vi+vf)time --> deltax/.5(vi+vf)=time
S: 125/.5(10.0+40.0)
S: 5.0 sec

that was the GUESS method. the G=givins, U=unkown, E=equation, S=solve, and S=solution.

deltax=the distance
vi=initial velocity
vf=final velocity

I hope this helped :)

Well, buckle up, because we're about to embark on a humorous journey through physics!

First, let's start off with some background knowledge. Acceleration is defined as the change in velocity over time. In this case, we're dealing with a uniform acceleration, which means the change in velocity is constant.

Given:
Initial velocity (u) = +10.0 m/s
Final velocity (v) = +40.0 m/s
Distance (s) = 125 m

We need to find the time it took for the car to cover that distance. To do that, we can use the following equation of motion:

v = u + at

Where:
v = Final velocity
u = Initial velocity
a = Acceleration
t = Time

Here's where the fun starts! Since we're dealing with uniform acceleration, we can use the average velocity to simplify our equation. The average velocity is the sum of the initial and final velocity divided by 2.

Average velocity = (u + v) / 2

Now let's plug in the values and solve this equation:

Average velocity = (10.0 m/s + 40.0 m/s) / 2
Average velocity = 50.0 m/s / 2
Average velocity = 25.0 m/s

Since acceleration is a constant, we can now calculate it using the equation:

a = (v - u) / t

Let's plug in again:

25.0 m/s = (40.0 m/s - 10.0 m/s) / t

Simplifying, we find:

25.0 m/s = 30.0 m/s / t

Now, let's isolate the variable t by cross-multiplying:

25.0 m/s * t = 30.0 m/s

Finally, divide both sides by 25.0 m/s to solve for t:

t = 30.0 m/s / 25.0 m/s
t = 1.2 s

So, after all the mathematical clowning around, the car took approximately 1.2 seconds to cover a distance of 125 meters. And remember, always drive with a smile!

To solve this problem, we can use the equation for uniformly accelerated motion:

vf = vi + at

Where:
vf is the final velocity (40.0 m/s),
vi is the initial velocity (10.0 m/s),
a is the acceleration (unknown),
t is the time (unknown).

We can also use the equation for displacement:

d = vit + 0.5at^2

Where:
d is the displacement (125 m).

Since we want to find the time, we can rearrange the first equation to isolate t:

t = (vf - vi) / a

Substituting the given values:

40.0 m/s - 10.0 m/s = a * t

30.0 m/s = a * t (Equation 1)

Now, let's use the second equation to express a in terms of d and t:

d = vi * t + 0.5 * a * t^2

125 m = 10.0 m/s * t + 0.5 * a * t^2

Since we already know that a = (vf - vi) / t from Equation 1, we can substitute it into the equation above:

125 m = 10.0 m/s * t + 0.5 * [(vf - vi) / t] * t^2

Simplifying:

125 m = 10.0 m/s * t + 0.5 * (30.0 m/s) * t

125 m = 10.0 m/s * t + 15.0 m/s * t

125 m = 25.0 m/s * t

Dividing both sides by 25.0 m/s:

t = 125 m / 25.0 m/s
t = 5.00 s

Therefore, it took 5.00 seconds to travel the distance of 125 meters.

To find the time it took for the car to go that distance, we can use the following equation:

v^2 = u^2 + 2as

Where:
v = final velocity (40.0 m/s)
u = initial velocity (10.0 m/s)
a = acceleration (uniform, so we can find it by dividing the change in velocity by the distance)
s = distance (125 m)

First, let's find the acceleration:
Δv = v - u = 40.0 m/s - 10.0 m/s = 30.0 m/s
a = Δv / s = 30.0 m/s / 125 m = 0.240 m/s^2

Now, we can rearrange the equation to solve for time (t):

v^2 = u^2 + 2as

40.0 m/s)^2 = (10.0 m/s)^2 + 2 * (0.240 m/s^2) * s

1600 m^2/s^2 = 100 m^2/s^2 + 0.480 m/s^2 * s

1500 m^2/s^2 = 0.480 m/s^2 * s

s = 1500 m^2/s^2 / 0.480 m/s^2

s = 3125 m

Now, we can solve for time:

s = 0.5 at^2 + ut

3125 m = 0.5 * 0.240 m/s^2 * t^2 + 10.0 m/s * t

0.120 m/s^2 * t^2 + 10.0 m/s * t - 3125 m = 0

This is a quadratic equation, and to solve it, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 0.120 m/s^2, b = 10.0 m/s, and c = -3125 m.

Plugging in these values, we get:

t = (-10.0 m/s ± √((10.0 m/s)^2 - 4 * 0.120 m/s^2 * -3125 m)) / (2 * 0.120 m/s^2)

t = (-10.0 m/s ± √(100.0 m^2/s^2 + 1500 m^2/s^2)) / 0.240 m/s^2

t = (-10.0 m/s ± √(1600 m^2/s^2)) / 0.240 m/s^2

t = (-10.0 m/s ± 40.0 m/s) / 0.240 m/s^2

Therefore, t = (-10.0 m/s + 40.0 m/s) / 0.240 m/s^2 or t = (-10.0 m/s - 40.0 m/s) / 0.240 m/s^2

Now, calculate:

t = (30.0 m/s) / 0.240 m/s^2 or t = (-50.0 m/s) / 0.240 m/s^2

t = 125 s or t = -208.3 s

Since time cannot be negative, the time it took for the car to go that distance is 125 s (to three significant figures).