# Acceleration Vector of a car making a turn without a change in speed

Acceleration Vector (2D) Vector = (Magnitude, Direction)

I am trying to calculate the acceleration vector acting on a car moving at a constant speed but changing direction. Example: A car is traveling 60 mph north (88 fps north (V1)). The car comes to a constant curve in the road that it takes 15 seconds to complete. The car’s velocity is now 88 fps east (V2). No change in speed, just change in direction.

It appears that the acceleration is dependent on the speed and time.

I have written a simulation and iteratively concluded that 9.2155 ft/sec^2 is the answer for my example.

I have: • V1 = Vector (88 fps, 90 degrees) • V2 = Vector (88 fps, 0 degrees) • Change in direction 90 degrees • Delta time 15 seconds • Degrees change per second = 6 • The acceleration is acting perpendicular to the velocity vector

What I have not figured out: • A formula to calculate the acceleration given the speed, degrees change in direction, and seconds to complete the change/turn.

The basic formulas are that $a=\frac {v^2}r$ and the time to complete a full circle is $t=\frac {2\pi r}v$ You can use these to get whatever you want. So if you take $t$ seconds to complete a quarter turn at speed $v$, your cover a linear distance of $vt$, so the circumference is $4vt$, the radius is $\frac{4vt}{2\pi}$ and the acceleration is $a=\frac {2\pi v^2}{4vt}=\frac {\pi v}{2t}$