Car traveling on circular track. Point A is at positive x,y. Point B is at positive x, negative y.
A car is traveling around a horizontal circular track with radius r = 220 m as shown. It takes the car t = 51 s to go around the track once. The angle èA = 31° above the x axis, and the angle èB = 56° below the x axis.
1)
What is the magnitude of the car’s acceleration?
2*3.14*220 = d
v= d/t ( put d and t=51 )
a = v^2/r ( r 220 ).
To find the magnitude of the car's acceleration, we need to use the formula for centripetal acceleration:
ac = (v^2) / r
Where:
- ac is the centripetal acceleration
- v is the velocity of the car
- r is the radius of the circular track
First, let's find the velocity of the car. Since we know the time taken to go around the track (t = 51 s) and the circumference of the circular track (2πr), we can calculate the average speed:
v = (2πr) / t
Now we can substitute the value of v into the centripetal acceleration formula:
ac = ((2πr) / t)^2 / r
Simplifying the equation gives us:
ac = (4π^2r) / t^2
Now we can plug in the values: r = 220 m and t = 51 s:
ac = (4π^2(220)) / (51)^2
Calculating the expression gives us the magnitude of the car's acceleration.