Coasting due west on your bicycle at 8.0 m/s, you encounter a sandy patch of road 7.2 m across. When you leave the sandy patch your speed has been reduced to 6.5 m/s. Assuming the bicycle slows with constant acceleration, what was its acceleration in the sandy patch?
average velocity (v) in the sandy patch ... (8.0 m/s + 6.5 m/s) / 2
time (t) to cross the patch ... 7.2 m / v
acceleration ... (6.5 m/s - 8.0 m/s) / t
Well, it seems like that sandy patch gave your bicycle a bit of a workout! To find the acceleration in the sandy patch, we can use the equation:
v² = u² + 2as
Where:
v is the final velocity (6.5 m/s)
u is the initial velocity (8.0 m/s)
a is the acceleration (what we're looking for)
s is the distance traveled across the sandy patch (7.2 m)
Rearranging the equation to solve for acceleration, we have:
a = (v² - u²) / (2s)
Let's plug in the values and calculate the acceleration, shall we?
a = (6.5² - 8.0²) / (2 * 7.2)
a = (42.25 - 64) / 14.4
a = -21.75 / 14.4
Drumroll, please...
a ≈ -1.513 m/s²
So, it seems like your bicycle was feeling a bit down in the sandy patch and experienced an acceleration of approximately -1.513 m/s². Keep an eye out for those sandy patches in the future, they can be quite the drag!
To find the acceleration of the bicycle in the sandy patch, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
- v is the final velocity (6.5 m/s)
- u is the initial velocity (8.0 m/s)
- a is the acceleration in the sandy patch (unknown)
- s is the distance covered in the sandy patch (7.2 m)
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (6.5^2 - 8.0^2) / (2 * 7.2)
Calculating:
a = (42.25 - 64) / 14.4
a = -21.75 / 14.4
a ≈ -1.51 m/s²
Therefore, the acceleration of the bicycle in the sandy patch was approximately -1.51 m/s². Note that the negative sign indicates that the bicycle decelerated during this time.
To find the acceleration, you can use the equation for constant acceleration:
v^2 = u^2 + 2as
where:
v is the final velocity (6.5 m/s),
u is the initial velocity (8.0 m/s),
a is the acceleration (what we want to find), and
s is the displacement (7.2 m).
First, let's substitute the known values into the equation:
(6.5 m/s)^2 = (8.0 m/s)^2 + 2a(7.2 m)
Simplifying this equation:
42.25 m^2/s^2 = 64.0 m^2/s^2 + 14.4a m
Rearranging the equation to solve for the acceleration:
14.4a = 42.25 m^2/s^2 - 64.0 m^2/s^2
14.4a = -21.75 m^2/s^2
Now, divide both sides by 14.4:
a = (-21.75 m^2/s^2) / 14.4
a ≈ -1.51 m/s^2
The negative sign in the result indicates that the bicycle's acceleration is in the opposite direction of its motion. Therefore, the acceleration in the sandy patch was approximately -1.51 m/s^2.