A train slows down at a constant rate as it
rounds a sharp circular horizontal turn. Its
initial speed is not known. It takes 19.8 s to
slow down from 81 km/h to 37 km/h. The
radius of the curve is 164 m.
As the train goes around the turn, what is
the magnitude of the tangential component of
the acceleration?
Answer in units of m/s2.
At the moment the train’s speed is 55 km/h,
what is the magnitude of the total accelera-
tion?
Answer in units of m/s2.
To find the magnitude of the tangential component of the acceleration, we can use the centripetal acceleration formula:
a_c = v^2 / r
where a_c is the centripetal acceleration, v is the velocity, and r is the radius of the circular turn.
First, let's convert the given speeds from km/h to m/s:
81 km/h = (81 * 1000 m) / (60 * 60 s) ≈ 22.5 m/s
37 km/h = (37 * 1000 m) / (60 * 60 s) ≈ 10.3 m/s
Now, we can calculate the centripetal acceleration at these speeds:
a_c1 = (22.5 m/s)^2 / 164 m ≈ 3.08 m/s^2
a_c2 = (10.3 m/s)^2 / 164 m ≈ 0.65 m/s^2
The tangential component of the acceleration is the rate at which the train's speed changes. Since the train is slowing down, the magnitude of the tangential component of acceleration is given by:
a_t = Δv / Δt
where a_t is the tangential component of acceleration, Δv is the change in velocity, and Δt is the time taken for the velocity to change.
We are given that the time taken to slow down from 81 km/h to 37 km/h is 19.8 seconds.
Δv = (37 m/s) - (22.5 m/s) ≈ 14.5 m/s
Δt = 19.8 s
Calculating the tangential component of acceleration:
a_t = (14.5 m/s) / (19.8 s) ≈ 0.73 m/s^2
Therefore, the magnitude of the tangential component of acceleration is 0.73 m/s^2.
To find the magnitude of the total acceleration when the train's speed is 55 km/h, we need to consider both the tangential and centripetal components of acceleration. Total acceleration can be found using the vector sum of the tangential and centripetal accelerations. Since they are at right angles to each other, we can use the Pythagorean theorem:
a_total = sqrt((a_c)^2 + (a_t)^2)
Substituting the values we calculated:
a_total = sqrt((3.08 m/s^2)^2 + (0.73 m/s^2)^2) ≈ 3.17 m/s^2
Therefore, the magnitude of the total acceleration when the train's speed is 55 km/h is approximately 3.17 m/s^2.