Assume the brakes of an automobile produce a constant deceleration of k ft/sec^2. Determine what k must be to bring an automobile traveling 60 mph to a rest in a distance of 100 ft from the point where breaks are applied?
With the same k, how far would a car traveling 30mph travel before being brought to a stop?
60mph = 88ft/s, so you want k where
88-kt=0 so t = 88/k
88t-1/2 kt^2 = 100
88(88/k) - 1/2 k (88/k)^2 = 100
k = 38.72 ft/s^2
now you can solve the other part.
Well, well, well, let's put on our math helmets and calculate some deceleration, shall we?
First things first, we need to convert that sneaky 60 mph into good old ft/sec. So buckle up, and let the conversion begin:
60 mph = 88 ft/sec (approximately)
Now, to determine the deceleration, we use the equation:
v^2 = u^2 + 2as
Where:
v is the final velocity (which is 0 because we want the car to come to a stop)
u is the initial velocity (88 ft/sec)
a is the deceleration (k ft/sec^2)
s is the distance (100 ft)
Since the car comes to a rest, we plug in the values and solve for k:
0 = (88 ft/sec)^2 + 2(k ft/sec^2)(100 ft)
Simplify, simplify, simplify...
0 = 7744 ft^2/sec^2 + 200k ft^2/sec^2
Now, my friend, we isolate k:
k = -7744 ft^2/sec^2 / 200 ft
Drumroll, please...
k = -38.72 ft/sec^2 (approximately)
So, to bring the car traveling at 60 mph to a stop in a distance of 100 ft, the brakes need to produce a deceleration of approximately -38.72 ft/sec^2. Negative? Yep, that's because it's deceleration. Brakes be like, "Slow down, my friend!"
Now, let's continue our entertaining journey and see how far a car traveling at 30 mph (which is 44 ft/sec) would travel before coming to a halt with the same deceleration:
Again, using the same equation with the new initial velocity:
0 = (44 ft/sec)^2 + 2(-38.72 ft/sec^2)(s ft)
More simplification magic...
0 = 1936 ft^2/sec^2 - 77.44s ft^2/sec^2
Now, it's isolation time for our beloved 's':
s = 1936 ft^2/sec^2 / 77.44 ft^2/sec^2
Grab your mathematic wigs and calculate...
s = 25 ft (approximately)
Hence, a car traveling at 30 mph (44 ft/sec) would travel approximately 25 ft before coming to a stop with the same deceleration. So remember, my friend, the faster you go, the longer the skid marks will be!
To determine the value of \(k\), we can use the kinematic equation that relates the final velocity (\(v\)), initial velocity (\(u\)), acceleration (\(a\)), and distance traveled (\(s\)):
\(v^2 = u^2 + 2as\)
Given that the initial velocity (\(u\)) is 60 mph and the distance traveled (\(s\)) is 100 ft, we need to convert these values to the appropriate units. The initial velocity (\(u\)) can be converted to ft/sec, and the distance traveled (\(s\)) should be in feet:
To convert 60 mph to ft/sec:
\(60 \text{ mph} = \frac{{60 \times 5280}}{{3600}} \text{ ft/sec} = 88 \text{ ft/sec}\)
Substituting these values into the equation, we have:
\(0 = (88)^2 + 2 \times k \times 100\)
Simplifying the equation:
\(0 = 7744 + 200k\)
To solve for \(k\), we can isolate it:
\(200k = -7744\)
\(k = -\frac{{7744}}{{200}}\)
\(k = -38.72\) ft/sec^2
Therefore, the value of \(k\) must be -38.72 ft/sec^2 to bring the automobile traveling at 60 mph to a stop in a distance of 100 ft.
Now, with the same \(k\), let's calculate how far a car traveling at 30 mph would travel before being brought to a stop.
To convert 30 mph to ft/sec:
\(30 \text{ mph} = \frac{{30 \times 5280}}{{3600}} \text{ ft/sec} = 44 \text{ ft/sec}\)
Using the same equation as before, substituting the values:
\(0 = (44)^2 + 2 \times k \times s\)
Simplifying the equation:
\(0 = 1936 + 88k\)
To solve for \(s\), we can isolate it:
\(88k = -1936\)
\(k = -\frac{{1936}}{{88}}\)
\(k = -22\) ft/sec^2
Therefore, with the same \(k\), a car traveling at 30 mph would travel \(s\) feet before being brought to a stop.
To determine the value of k, we can use the equations of motion. The first equation we can use is:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
First, let's convert the initial velocity from 60 mph to ft/sec. We know that 1 mph is equal to 1.46667 ft/sec, so:
u = 60 mph * 1.46667 ft/sec/mph
u = 88 ft/sec
Next, we need to find the final velocity (v). Since the car comes to a rest, v will be zero.
The displacement (s) is given as 100 ft.
Plugging these values into the equation, we have:
0^2 = (88 ft/sec)^2 + 2 * k * 100 ft
This equation simplifies to:
0 = 7744 + 200k
Now, we can solve for k:
200k = -7744
k = -7744 / 200
k = -38.72 ft/sec^2
Therefore, k must be approximately -38.72 ft/sec^2 to bring the car from 60 mph to rest in a distance of 100 ft.
Now, let's find out how far the car would travel before being brought to a stop if it was initially traveling at 30 mph.
Using the same value of k, we can use the equations of motion again. We need to find the distance (s). The initial velocity (u) is given as 30 mph, which is equal to 44 ft/sec.
The final velocity (v) will still be zero.
Plugging these values into the equation v^2 = u^2 + 2as, we have:
0^2 = (44 ft/sec)^2 + 2 * k * s
0 = 1936 + 2ks
Since k remains the same, we can rearrange the equation to solve for s:
s = -1936 / (2k)
s = -1936 / (2 * -38.72)
s = 24.99 ft
Therefore, with the same value of k, the car traveling at 30 mph would travel approximately 24.99 ft before being brought to a stop.