A train traveling at a speed of 110 km/h uniformly decelerates for 20 seconds. During that time it travels 450 m.
a. What is the acceleration ? = m/s2
b. What is its speed after the deceleration ? = km/h
s = v0*t + 1/2 at^2
plug in your numbers for v0 (convert to m/s) and t.
now, use that value of a for
v = v0 + at
and then convert m/s to km/hr
To find the acceleration of the train, we can use the formula:
acceleration = (final velocity - initial velocity) / time
Given:
Initial velocity (u) = 110 km/h
Time (t) = 20 seconds
a. First, we need to convert the initial velocity from km/h to m/s:
To convert km/h to m/s, divide by 3.6 (since 1 km/h = 1/3.6 m/s):
initial velocity in m/s (u) = 110 km/h / 3.6 = 30.56 m/s
Next, we need to find the final velocity of the train. We know that the train is uniformly decelerating, which means its final velocity will be less than its initial velocity.
We can find the final velocity using the equation:
final velocity = initial velocity + (acceleration * time)
However, since the train is decelerating, the acceleration is negative.
Let's assume the final velocity is v.
Given:
initial velocity (u) = 30.56 m/s
time (t) = 20 seconds
final velocity (v) = ? (to be determined)
v = u + (acceleration * t)
v = 30.56 + (acceleration * 20)
Now, we also know that the distance traveled during the deceleration is 450 m.
We can use the equation:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Substituting the known values:
450 = (30.56 * 20) + (0.5 * acceleration * 20^2)
450 = 611.2 + 200 * acceleration
Rearranging the equation:
200 * acceleration = 450 - 611.2
200 * acceleration = -161.2
Now, we can solve for acceleration:
acceleration = (-161.2) / 200 = -0.806 m/s^2
b. To find the speed after deceleration, we can use the final velocity value obtained:
final velocity in km/h = v * 3.6
final velocity in km/h = (-0.806 m/s * 20 s) * 3.6 / 1000 = -0.579 km/h
The speed after deceleration is approximately -0.579 km/h. Note that the negative sign indicates that the train has slowed down.