a) A light-rail commuter train accelerates at a rate of 1.25 m/s2. How long does it take it to reach its top speed of 80.0 km/h starting from rest?
(b) The same train ordinarily decelerates at a rate of 1.75 m/s2. How long does it take to come to a stop from its top speed?
(c) In emergencies the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in m/s2?
V = 80km/h = 80000m/3600s = 22.22 m/s.
a. V = Vo + a*t = 22.22 m/s.
0 + 1.25t = 22.22. Solve for t.
b. V = Vo + a*t = 0
22.22 - 1.75*t = 0. Solve for t.
c. V = Vo + a*t = 0
22.22 + a*8.30 = 0. Solve for a; it
should be negative.
17.776
12
V = 80km/h = 80000m/3600s = 22.22 m/s.
a. V = Vo + a*t = 22.22 m/s.
0 + 1.25t = 22.22. Solve for t.
b. V = Vo + a*t = 0
22.22 - 1.75*t = 0. Solve for t.
c. V = Vo + a*t = 0
22.22 + a*8.30 = 0. Solve for a; it
should be negative
To solve these questions, we need to use the equations of motion related to acceleration, velocity, and time. These equations are:
1) v = u + at - relates final velocity (v), initial velocity (u), acceleration (a), and time (t)
2) s = ut + (1/2)at^2 - relates displacement (s), initial velocity (u), acceleration (a), and time (t)
3) v^2 = u^2 + 2as - relates final velocity (v), initial velocity (u), acceleration (a), and displacement (s)
Now let's solve each question step-by-step:
(a) To find the time to reach top speed, we need to use Equation 1:
Given:
Initial velocity (u) = 0 m/s (starting from rest)
Acceleration (a) = 1.25 m/s^2
Top speed (v) = 80.0 km/h = 80.0 * 1000 / 3600 = 22.22 m/s
v = u + at
22.22 = 0 + 1.25t
Solving for t, we get:
t = (22.22 - 0) / 1.25
t = 17.776 s
Therefore, it takes 17.776 seconds for the train to reach its top speed.
(b) To find the time to come to a stop, we need to use Equation 1:
Given:
Initial velocity (u) = 80.0 km/h = 80.0 * 1000 / 3600 = 22.22 m/s
Acceleration (a) = -1.75 m/s^2 (negative sign because it is decelerating)
Final velocity (v) = 0 m/s (coming to a stop)
v = u + at
0 = 22.22 + (-1.75)t
Solving for t, we get:
t = (0 - 22.22) / (-1.75)
t = 12.688 s
Therefore, it takes 12.688 seconds for the train to come to a stop from its top speed.
(c) To find the emergency deceleration, we need to use Equation 2:
Given:
Initial velocity (u) = 80.0 km/h = 80.0 * 1000 / 3600 = 22.22 m/s
Time (t) = 8.30 s
Displacement (s) = ? (we don't have this value, so we will solve for it to find the emergency deceleration)
s = ut + (1/2)at^2
s = 22.22 * 8.30 + (1/2) * a * (8.30)^2
Solving for a, we rearrange the equation as follows:
a = (2s - 22.22 * 8.30) / (8.30)^2
By substituting the given values, we get:
a = (2 * s - 22.22 * 8.30) / (8.30)^2
Therefore, the emergency deceleration in m/s^2 can be calculated using the given equation once the displacement (s) value is known.