A car is slowing down at a rate of 20 km/h/s. How far does it travel if its original velocity is 50 km/h and its final velocity is 5.0 m/s?
thank you
To find the distance traveled by the car, we need to use the equation of motion:
\(v^2 = u^2 + 2as\),
where:
\(v\) = final velocity (in m/s),
\(u\) = initial velocity (in m/s),
\(a\) = acceleration (in m/s^2),
\(s\) = distance (in meters).
First, let's convert the given velocities from km/h to m/s:
\(50 \, \text{km/h} = \frac{50 \times 1000}{3600} \, \text{m/s}\),
\(5.0 \, \text{m/s} = 5.0 \, \text{m/s}\).
Now we can solve for the acceleration (a):
\(\text{change in velocity} = \text{final velocity} - \text{initial velocity}\),
\(\Delta v = v - u\),
\(\Delta v = 5.0 - \frac{50 \times 1000}{3600}\).
Using the equation \(a = \frac{\Delta v}{t}\), we can find the acceleration:
\(a = \frac{\Delta v}{t}\),
\(a = \frac{5.0 - \frac{50 \times 1000}{3600}}{t}\),
\(a = \frac{5.0 \times 3600 - 50 \times 1000}{3600t}\),
\(a = \frac{-1000}{t}\).
Given that the acceleration is \(-20 \, \text{m/s}^2\) (since the car is slowing down), we can solve for \(t\):
\(-20 = \frac{-1000}{t}\),
\(t = \frac{-1000}{-20}\),
\(t = 50 \, \text{s}\).
Now, we can use the equation \(v = u + at\) to find the distance:
\(v = u + at\),
\(5.0 = \frac{50 \times 1000}{3600} + (-20) \times 50\),
\(5.0 = \frac{50000}{3600} - 1000\),
\(5.0 = 13.89 - 1000\),
\(5.0 + 985 = 13.89\),
\(s = 18.89 \, \text{m}\).
Therefore, the car travels a distance of 18.89 meters.
To find the distance traveled by the car, we need to use the equation of motion that relates the initial velocity, final velocity, acceleration, and distance traveled.
The equation we will use is:
v^2 = u^2 + 2as
Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- s is the distance traveled
First, let's convert the initial velocity and final velocity to the same unit. Since the final velocity is given in m/s, let's convert the initial velocity to m/s as well.
Given:
Initial velocity, u = 50 km/h
Final velocity, v = 5.0 m/s
Converting the initial velocity to m/s:
u = 50 km/h = (50 * 1000) / 3600 = 13.89 m/s (approximately)
Now we have all the values we need to substitute into the equation.
v^2 = u^2 + 2as
Substituting the given values:
(5.0)^2 = (13.89)^2 + 2 * 20 * s
25.0 = 193.21 + 40s
Now, let's solve for s.
25.0 - 193.21 = 40s
-168.21 = 40s
Dividing both sides by 40:
s = -168.21 / 40
s ≈ -4.20525 meters
Since distance cannot be negative, we can conclude that the car has traveled approximately 4.20525 meters.
Change km to m, h to sec
20km/h/s= 20000/3600 m/s^2
50km/h= 50000/3600 m/s
Vf^2=Vi^2+2ad where a= -20/3.6 m/s^2
solve for d.
Well, let me put on my clown nose and tackle this question. So, the car is slowing down at a rate of 20 km/h/s. That's like molasses in winter – pretty slow!
Now, let's get down to business. We know the car's original velocity is 50 km/h and its final velocity is 5.0 m/s.
To find the distance it travels, we need to calculate the time it takes for the car to slow down from 50 km/h to 5.0 m/s.
First, let's convert everything to the same units. Since we want the final velocity in km/h, let's convert the 5.0 m/s to km/h.
Converting 5.0 m/s to km/h is like trying to catch a clown with sticky shoes – it takes some effort. It's around 0.018 km/h.
To find the time it takes, we use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, v = 0.018 km/h, u = 50 km/h, and a = -20 km/h/s (negative because the car's slowing down).
Now, we rearrange the formula to solve for t: t = (v - u) / a.
Let's plug in the numbers: t = (0.018 - 50) / -20.
Calculating that, we get a time of... *drumroll*... 2.5009 hours.
Finally, to find the distance, we use the formula s = ut + 0.5at^2.
Since the car starts at rest (0 km/h), we can simplify our formula to s = 0.5at^2.
Now, we plug in the numbers: s = 0.5 * -20 * (2.5009)^2.
Crunching through the numbers, the distance the car travels is approximately 15.63 kilometers.
So, after all that clowning around with formulas, the car travels approximately 15.63 kilometers. Drive safely, my friend!