Question

A race car accelerated uniformly from a speed of 40 m/s to a speed of 58 m/s in 6 seconds while traveling around a circular track of radius 528 m. When the magnitude of the car's total acceleration becomes 5 m/s2. what is its speed (in m/s)? Group of answer choices

Answer 1

Answer: 45.95 m/s

Step-by-step explanation:

When we talk about circular motion, the object's acceleration
`a` (which is a vector quantity) has two components: the centripetal acceleration
`a_(C)` always directed to the center of the circular track and the tangential acceleration
`a_(T)` which is tangent to the circular path.

Since both vectors are perpendicular to each other, the magnitude of
`a` can be calculated by the Pithagorean Theorem:

`a^(2)=a_(C)^(2) + a_(T)^(2)` (1)

Where:

`a=5 m/s^(2)`

`a_(C)=(V^(2))/(r)` where
`V` is the speed and
`r=528 m` is the radius of the circle

`a_(T)` can be calculated knowing the initial speed (
`V_(o)=40 m/s`) and final speed (
`V_(f)=58 m/s`) of the car and the time (
`t=6 s`) it takes to accelerate at this constant rate:

`a_(T)=(V_(f) - V_(o))/(t)`

`a_(T)=(58 m/s - 40 m/s)/(6 s)=3m/s^(2)`

Rewritting (1):

`a^(2)=((V^(2))/(r))^(2) + a_(T)^(2)` (2)

Isolating
`V`:

`V=((a^(2) - a_(T)^(2))r^(2))^(1/4)` (3)

`V=(((5 m/s^(2))^(2) - (3 m/s^(2))^(2))(528 m)^(2))^(1/4)` (4)

Finally:

`V=45.956 m/s`