An electric train starting from rest is uniformly accelerated in a distance of 40m to a velocity of 10m/s, for this journey.
(I).Calculate the time taken by the train.
(II). The acceleration of the train.
vf^2=vi^2+2ad solve for acceleration a, vi=0
knowing acceleration a
d=1/2 a*time^2
time=sqrt(2*distance/acceleration)
To answer these questions, we can use the kinematic equation:
v² = u² + 2as
where
v = final velocity
u = initial velocity
a = acceleration
s = displacement
Given:
u = 0 m/s (train starts from rest)
v = 10 m/s
s = 40 m
(I). To calculate the time taken by the train, we need to rearrange the equation to solve for time (t):
v = u + at
Using the given values, the equation becomes:
10 = 0 + a * t
Simplifying, we have:
10 = a * t
Therefore, t = 10 / a
(II). To calculate the acceleration of the train, we can rearrange the first equation to solve for a:
v² = u² + 2as
10² = 0² + 2 * a * 40
100 = 80a
Divide both sides by 80:
a = 100 / 80
a = 1.25 m/s²
Therefore, the time taken by the train is t = 10 / a, and the acceleration of the train is a = 1.25 m/s².
To calculate the time taken by the train, we can use the formula:
\(v = u + at\)
Where:
\(v\) is the final velocity of the train,
\(u\) is the initial velocity of the train (which is 0 as it starts from rest),
\(a\) is the acceleration of the train, and
\(t\) is the time taken.
Let's plug in the given values:
\(v = 10 \, \text{m/s}\)
\(u = 0 \, \text{m/s}\)
\(a = ?\)
\(t = ?\)
From the given information, we know that the train starts from rest and accelerates uniformly over a distance of 40m to a velocity of 10m/s. We do not have the time, but we can find it using the formula:
\(s = ut + \frac{1}{2}at^2\)
Where:
\(s\) is the distance traveled by the train,
\(u\) is the initial velocity of the train,
\(a\) is the acceleration of the train, and
\(t\) is the time taken.
Let's plug in the given values:
\(s = 40 \, \text{m}\)
\(u = 0 \, \text{m/s}\)
\(a = ?\)
\(t = ?\)
We want to find the time, so let's rearrange the formula:
\(\frac{1}{2}at^2 + ut - s = 0\)
Now we can use the quadratic formula to solve for time:
\(t = \frac{-u \pm \sqrt{u^2 - 4(\frac{1}{2}a)(-s)}}{2(\frac{1}{2}a)}\)
Since we know the train starts from rest (\(u = 0\)), the formula becomes:
\(t = \frac{\sqrt{0 - 4(\frac{1}{2}a)(-s)}}{a}\)
Simplifying further:
\(t = \frac{2 \sqrt{as}}{a}\)
Now we can solve for \(t\) using the given values:
\(t = \frac{2 \sqrt{a(40)}}{a}\)
Simplifying:
\(t = \frac{2 \sqrt{40a}}{a}\)
So, the time taken by the train can be found using the formula \(t = \frac{2 \sqrt{40a}}{a}\).
Now let's move on to finding the acceleration of the train.
We already have the formula:
\(v = u + at\)
Plugging in the given values:
\(v = 10 \, \text{m/s}\)
\(u = 0 \, \text{m/s}\)
\(a = ?\)
\(t = ?\)
From the given information, we can see that the train accelerates uniformly over a distance of 40m to a velocity of 10m/s. We do not have the acceleration, but we can find it using the formula:
\(v^2 = u^2 + 2as\)
Let's plug in the given values:
\(v = 10 \, \text{m/s}\)
\(u = 0 \, \text{m/s}\)
\(a = ?\)
\(s = 40 \, \text{m}\)
Rearranging the formula, we get:
\(a = \frac{v^2 - u^2}{2s}\)
Substituting the given values:
\(a = \frac{10^2 - 0^2}{2 \times 40}\)
Simplifying:
\(a = \frac{100}{80}\)
So, the acceleration of the train is \(a = \frac{5}{4} \, \text{m/s}^2\).
To summarize:
(I). The time taken by the train can be found using the formula \(t = \frac{2 \sqrt{40a}}{a}\).
(II). The acceleration of the train is \(a = \frac{5}{4} \, \text{m/s}^2\).