A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a velocity of 42 m/s. What is the approximate acceleration of the train during this time?
v^2 = 2as, so
42^2 = 2a*5600
a = 0.1575 m/s^2
To find the approximate acceleration of the train, we can use the equation for uniform acceleration:
\(v^2 = u^2 + 2as\)
Where:
\(v\) = final velocity (42 m/s)
\(u\) = initial velocity (0 m/s, since the train starts from rest)
\(a\) = acceleration (unknown)
\(s\) = distance traveled (5.6 km = 5600 m)
First, let's convert the distance to meters:
\(s = 5.6 \times 1000 = 5600 \, \text{m}\)
Now, we can plug in these values into the equation and solve for acceleration:
\(42^2 = 0^2 + 2a \times 5600\)
Simplifying the equation gives:
\(1764 = 2a \times 5600\)
Divide both sides of the equation by 2 and 5600:
\(a = \frac{1764}{2 \times 5600}\)
\(a \approx \frac{1764}{11200}\)
\(a \approx 0.157 \, \text{m/s}^2\)
Therefore, the approximate acceleration of the train during this time is approximately \(0.157 \, \text{m/s}^2\).