Question

A car travels at a constant speed around a circular track whose radius is 2.77 km. The car goes once around the track in 361 s. What is the magnitude of the centripetal acceleration of the car? A car travels at a constant speed around a circular track whose radius is 2.77 km. The car goes once around the track in 361 s. What is the magnitude of the centripetal acceleration of the car? @Physics

Answer 1

Centripetal acceleration is given by:
\[\Large {a_c} = \frac{{{v^2}}}{r}\]
To get the velocity, work out the linear distance of the track - i.e. the circumference of the circle - and how far was travelled in this time, remembering to convert from km to m (1km = 1*10^3m):
\[\Large d = 2\pi r = 2 \cdot \pi \cdot 2.77 \times {10^3} = 17404{\rm{ m}}\]
Given that the velocity was constant, we can then take the average velocity:
\[\Large v = \frac{d}{t} = \frac{{17404}}{{361}} = 48.2{\rm{ m}}{{\rm{s}}^{ - 1}}\]
Plug this into the centripetal acceleration formula:
\[\Large {a_c} = \frac{{{{48.2}^2}}}{{2.77 \times {{10}^3}}} = 0.839{\rm{ m}}{{\rm{s}}^{ - 2}}\]And there we have it.