The driver of a train traveling at 30 m/s applies the brakes when he passes an amber signal. The next signal is 1.5 km/h down the track and it reaches 75 seconds later. Assuming uniform acceleration, find the velocity of the train at the second signal.
v = 30 - a (75)
assume you mean 1.5 km or 1500 meters
1500 = 30 (75) - .5 a (75)^2
solve for a
then go back and get v at 75 seconds
-30 m/s backward
Well, if the train driver sees an amber signal, it's like the universe is telling them, "Hey, maybe you should slow down." But let's find the velocity of the train at the second signal using some good old physics humor.
So, we are given that the train's initial velocity is 30 m/s. And we know that the time it takes to reach the second signal is 75 seconds. The distance between the two signals is 1.5 kilometers.
To start, we need to convert the distance to meters because physics likes to keep everything in SI units. There are 1000 meters in a kilometer, so 1.5 kilometers is 1500 meters. Now we're all speaking the same language.
We can use the equation of motion: distance = initial velocity * time + 0.5 * acceleration * time^2. Since we're assuming uniform acceleration, we can express it simply as distance = initial velocity * time.
So, with all the given information, we have:
1500 meters = 30 m/s * 75 seconds.
Now, let's solve for velocity at the second signal. Dividing both sides of the equation by 75 seconds, we get:
20 meters/second = 30 m/s.
Wait, what's going on? It seems like the velocity at the second signal is actually less than the initial velocity. But fear not, my young question-wielder! This is because the train is slowing down due to the application of brakes.
So, using a bit of humor and some physics, we can conclude that the velocity of the train at the second signal is 20 m/s.
To solve this problem, we can use the equation of motion that relates distance, initial velocity, final velocity, time, and acceleration.
The equation we will use is:
\[d = ut + \frac{1}{2}at^2\]
Where:
d is the distance traveled,
u is the initial velocity,
t is the time taken,
and a is the acceleration.
In this case, the train is initially traveling at 30 m/s, and the time taken to reach the second signal is given as 75 seconds. The distance between the signals is 1.5 km, which is 1500 meters.
Now, we need to find the acceleration of the train. Assuming uniform acceleration, we can rearrange the equation as follows:
\[d = ut + \frac{1}{2}at^2\]
\[1500 = 30(75) + \frac{1}{2}a(75^2)\]
Simplifying:
\[1500 = 2250 + \frac{1}{2}a(5625)\]
\[1500 - 2250 = \frac{1}{2}a(5625)\]
\[-750 = \frac{1}{2}a(5625)\]
Now, we can solve for acceleration (a) by dividing both sides by 5625 and then multiplying by 2:
\[-750 = \frac{1}{2}a(5625)\]
\[-\frac{750}{5625} = \frac{1}{2}a\]
\[-\frac{1}{7.5} = \frac{1}{2}a\]
\[-\frac{1}{7.5} \times 2 = a\]
\[-\frac{2}{7.5} = a\]
\[-\frac{4}{15} = a\]
Now that we have the acceleration, we can use it to find the velocity at the second signal. We can use the equation:
\[v = u + at\]
Plugging in the values:
\[v = 30 + (-\frac{4}{15})(75)\]
\[v = 30 + (-\frac{4}{15})(75)\]
\[v = 30 - 20\]
\[v = 10 m/s\]
Therefore, the velocity of the train at the second signal is 10 m/s.