Question

A car enters a circular off ramp (radius of 180 m) at 30 m/s. The car's on - bond accelerometers sense a total acceleration magnitude of 7.07 m/s2. What is the car's acceleration magnitude a2 (the component tangent to its path)?

Answer 1

Horizontal acceleration will be
`2.07m/sec^2`

Step-by-step explanation:

We have given radius of track = 180 m

velocity = 30 m/sec

Total acceleration
`a_t=7.07m/sec^2`

The centripetal;l acceleration which is always normal to the velocity also known as normal acceleration is given by
`a_n=(v^2)/(r)=(30^2)/(180)=5m/sec^2`

So the tangent acceleration will be
`a_t=7.07-5=2.07m/sec^2`