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

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:

$a=5 m/s^{2}$

$a_{C}=\frac{V^{2}}{r}$ 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:

$a_{T}=\frac{V_{f} - V_{o}}{t}$

$a_{T}=\frac{58 m/s - 40 m/s}{6 s}=3m/s^{2}$

Rewritting (1):

$a^{2}=(\frac{V^{2}}{r})^{2} + a_{T}^{2}$ (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:

$V=45.956 m/s$