A train slows down as it rounds a sharp horizontal turn, going from 99.0 km/h to 41.0 km/h in the 13.5 s it takes to round the bend. The radius of the curve is 175 m. Compute the acceleration at the moment the train speed reaches 41.0 km/h. Assume the train continues to slow down at this time at the same rate.

change speeds to m/s

then compute the tangential acceleration, changevelocyt/time

now compute the centripetal acceleration
v^2/r

Because these are at right angles, then
accelearation= sqrt(tangential^2+centripetal^2)

To compute the acceleration at the moment the train speed reaches 41.0 km/h, we can use the kinematic equation:

v^2 = u^2 + 2as

Where:
- v is the final velocity (41.0 km/h),
- u is the initial velocity (99.0 km/h),
- a is the acceleration,
- s is the distance.

First, let's convert the velocities from kilometers per hour to meters per second:

v = 41.0 km/h * (1000 m / 1 km) * (1 h / 3600 s)
u = 99.0 km/h * (1000 m / 1 km) * (1 h / 3600 s)

v = 11.4 m/s
u = 27.5 m/s

Next, let's calculate the distance traveled by the train during the slowing down period:

d = radius * angle

To find the angle, we can use the equation:

v = u * (π / 180) * angle / t

Rearranging the equation:

angle = (v * t / u) * (180 / π)

angle = (11.4 m/s * 13.5 s / 27.5 m/s) * (180 / π)

angle ≈ 94.88 degrees

Now, let's calculate the distance:

d = 175 m * (94.88 degrees / 360 degrees)

d ≈ 45.95 m

Now, we can substitute the values into the kinematic equation:

v^2 = u^2 + 2as

(11.4 m/s)^2 = (27.5 m/s)^2 + 2 * a * 45.95 m

130.16 m^2/s^2 = 756.25 m^2/s^2 + 91.9 m * a

Rearranging the equation and solving for the acceleration:

a = (130.16 m^2/s^2 - 756.25 m^2/s^2) / (2 * 45.95 m)

a ≈ -8.07 m/s^2

Therefore, the acceleration at the moment the train speed reaches 41.0 km/h is approximately -8.07 m/s^2.