The firetruck goes around a 180°, 162 m radius circular curve. It enters the curve with a speed of 12.6 m/s and leaves the curve with a speed of 38.8 m/s. Assuming the speed changes at a constant rate, what is the magnitude of the total acceleration of the firetruck just after it has entered the curve?
From the length of the curve (L=πr), calculate the average tangential acceleration, At
= (38.8-12.6)/L
Find the radial acceleration due to curvature at entry to curve (v=12.6m/s)
Ar=v²/r
=12.6²/r
Since the two accelerations are perpendicular to each other, add Ar and At vectorially.
I don't understand how you get At = (38.8 - 12.6)/L.Please explain
Thanks.
To find the magnitude of the total acceleration of the firetruck just after it has entered the curve, we need to consider the centripetal acceleration and the tangential acceleration.
1. Centripetal acceleration (ac):
The centripetal acceleration is the acceleration towards the center of the circular path. It can be calculated using the formula:
ac = (v^2) / r
where v is the speed of the firetruck and r is the radius of the circular curve.
Given that the speed of the firetruck is 12.6 m/s and the radius of the curve is 162 m, we can calculate the centripetal acceleration:
ac = (12.6^2) / 162
ac ≈ 0.979 m/s^2
2. Tangential acceleration (at):
The tangential acceleration is the change in speed of the firetruck along the tangent to the circular path. It can be calculated using the formula:
at = (vf - vi) / t
where vf is the final speed of the firetruck, vi is the initial speed of the firetruck, and t is the time taken.
Given that the initial speed (vi) is 12.6 m/s, the final speed (vf) is 38.8 m/s, and we assume the speed changes at a constant rate, we can calculate the tangential acceleration:
at = (38.8 - 12.6) / t
3. Total acceleration (a):
The total acceleration is the vector sum of the centripetal acceleration (ac) and the tangential acceleration (at). Since they are perpendicular to each other, we can calculate the magnitude using the Pythagorean theorem:
a = sqrt(ac^2 + at^2)
Substituting the calculated values:
a = sqrt((0.979^2) + ((38.8 - 12.6) / t)^2)
Unfortunately, there is missing information regarding the time taken for the speed to change, denoted as 't'. Without this information, we cannot calculate the exact magnitude of the total acceleration of the firetruck just after it has entered the curve.