The key information for the problem where a car moves along a circular track of radius 50.0 m, starting from rest with a tangential acceleration of 3.00 m/s², and the situation where its centripetal acceleration equals its tangential acceleration is as follows:
- Centripetal acceleration ac=v2ra_c=\frac{v^2}{r}ac=rv2 where vvv is the tangential velocity and r=50.0r=50.0r=50.0 m.
- Tangential acceleration at=3.00 m/s2a_t=3.00,m/s^2at=3.00m/s2.
- Condition: ac=ata_c=a_tac=at.
From this,
- v250=3\frac{v^2}{50}=350v2=3
- v2=150v^2=150v2=150
- v=150≈12.25 m/sv=\sqrt{150}\approx 12.25,m/sv=150≈12.25m/s.
Using kinematic equations, the time ttt taken to reach this velocity with constant tangential acceleration from rest is:
- v=attv=a_ttv=att so t=vat=12.253≈4.08 st=\frac{v}{a_t}=\frac{12.25}{3}\approx 4.08,st=atv=312.25≈4.08s.
The total acceleration magnitude (since centripetal and tangential accelerations are perpendicular) is:
- a=ac2+at2=32+32=32≈4.24 m/s2a=\sqrt{a_c^2+a_t^2}=\sqrt{3^2+3^2}=3\sqrt{2}\approx 4.24,m/s^2a=ac2+at2=32+32=32≈4.24m/s2.
The angle θ\theta θ traveled around the track in radians when this occurs is 1 rad. The distance traveled by the car during this time is about 25 m. Thus, the important results when centripetal acceleration equals tangential acceleration for this problem are:
- Angle around the track traveled: 1 rad.
- Magnitude of total acceleration: 32 m/s23\sqrt{2},m/s^232m/s2.
- Time elapsed: 503≈4.08 s\sqrt{\frac{50}{3}}\approx 4.08,s350≈4.08s.
- Distance traveled: 25 m.
These outcomes align with the physics concept that the car accelerates tangentially while increasing its centripetal acceleration due to increasing speed on the circular track.
