A train slows down as it rounds a sharp horizontal turn, slowing from 87.4 km/h to 44.6 km/h in the 16.1 s that it takes to round the bend. The radius of the curve is 180 m. Compute the acceleration at the moment the train speed reaches 44.6 km/h. Assume that it continues to slow down at this time at the same rate. what or how do i find the magnitude and dierction?
magnitude acceleration=(87.4-44.6)/16.1 is the tangential acceleration, in the direction it is going.
Now there is another acceleartion, directed at the center of the circle, centripetal acceleration, 44.1^2/180
These two accelerations are at 90 degrees to each other.
Magnitude= sqrt(tangential magnitude^2+centripetal^2)
direction, theta=arctan centriptal/tangential where theta is the angle measured from the tangent towards the radius.
Where to get 44.1 for Velocity as in the Centripetal Acceleration?
To find the magnitude and direction of the acceleration at the moment the train speed reaches 44.6 km/h, we can use the centripetal acceleration formula.
The formula for centripetal acceleration is given by:
a = (v² / r)
Where:
a = centripetal acceleration
v = velocity of the train (44.6 km/h)
r = radius of the curve (180 m)
First, we need to convert the velocity from km/h to m/s, as the radius is given in meters.
1 km/h = 1000 m/3600 s = 0.2778 m/s
Now, let's convert the velocity to m/s:
v = 44.6 km/h * 0.2778 m/s
v = 12.39 m/s
Now, we can solve for the acceleration:
a = (v² / r)
a = (12.39 m/s)² / 180 m
Calculating the acceleration:
a = 153.60 m²/s² / 180 m
a = 0.8533 m/s²
So, the magnitude of the acceleration at the moment the train speed reaches 44.6 km/h is 0.8533 m/s².
Now, to determine the direction of the acceleration, we know that the acceleration always points towards the center of the curve. In this case, since the train is slowing down, the acceleration will be directed inward, in the opposite direction of the train's velocity vector.
Therefore, the direction of the acceleration is inward or towards the center of the curve.