A train slows down as it rounds a sharp bend, slowing from 90.0km/h to 50.0km/h in the 15.0s that it takes to round the bend. The radius of the curve is 150m.
Compute the acceleration at the moment the train speed reaches 50.0km/h. Assume it continues to slow down at the same rate.
The acceleration is the vector sum of a centripetal component and a tangential component (along the direction of motion). The two componets are perpendicular.
The centripetal component is
a1 = V^2/R
Make sure you convert the 50 km/hr to meters per second, to get the answer in m/s^2. 50,000m/3600 s = 13.89 m/s
The 40 km/h speed reduction is 11.11 m/s. Divide that by 15.0 s for the tangential acceleration in m/s^2.
How did you find the speed reduction?
The speed reeduction is 90 - 50 km/h. That is 40 km/h
Converted into m/s, I presume?
Converted to m/s, I presume?
To compute the acceleration at the moment the train speed reaches 50.0 km/h, we can use the formula for centripetal acceleration:
a = (v^2) / r
where
a = acceleration
v = velocity
r = radius of the curve
First, let's convert the velocities from km/h to m/s, since the formula requires the velocity to be in m/s:
v1_initial = 90.0 km/h = 90.0 * (1000 m / 3600 s) = 25.0 m/s
v2_final = 50.0 km/h = 50.0 * (1000 m / 3600 s) = 13.9 m/s
Now, let's calculate the initial centripetal acceleration using the initial velocity v1_initial and the radius r:
a_initial = (v1_initial^2) / r = (25.0 m/s)^2 / 150 m = 4.17 m/s^2
Since it is mentioned that the train continues to slow down at the same rate, the acceleration remains constant as the train slows down. Therefore, the acceleration when the train reaches 50.0 km/h is the same as the initial acceleration:
a_final = a_initial = 4.17 m/s^2
So, the acceleration at the moment the train speed reaches 50.0 km/h is 4.17 m/s^2.