Question

5.
A racing car is moving on a circular track at a
constant speed of 25 m/s. It takes 20 s to complete
one loop. The magnitude of centripetal acceleration
of the car will be approximately
(1) 5 m/s2
(2) 3 m/s2
(3) 8 m/s2
(4) 10 m/s2

Answer 1

$\huge\mathfrak\red{Answer}$

$total \: distance = 2 \: \pi \: r \\ where \: r \: is \: the \: radius \: of \: the \: \\ circular \: track \\ \\ speed(v) = 25m {sec}^{ - 1} \\ time \: taken = 20sec \\ \\ distance = speed \times time \\ 2 \: \pi \: r = 25 \times 20 \\ \\ r = \frac{20 \times 25}{2 \times 3.14} \\ \\ r = \frac{500}{6.28} \\ \\ r = 79.61m \\ \\ centripetal \: acceleration = \frac{ {v}^{2} }{r} \\ centripetal \: acceleration = \frac{25 \times 25}{79.61} \\ centripetal \: acceleration = \frac{625}{79.61} \\ centripetal \: acceleration = 7.8m {sec}^{ - 2} \\ \\ |answer| = 8m {sec}^{ - 2}$

Answer 2

The magnitude of centripetal acceleration  of the car will be approximately $8\ m/s^2$.

Explanation:

Speed of the car, v = 25 m/s

Time taken, t = 20 s

On a circular track, the speed of the car is given by :

$v=\dfrac{d}{t}=\dfrac{2\pi r}{t}$

$r=\dfrac{vt}{2\pi}$

$r=\dfrac{25\times 20}{2\pi}$

r = 79.57 m

The centripetal acceleration of the car is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(25\ m/s)^2}{79.57\ r}$

$a=7.85\ m/s^2$

or

$a=8\ m/s^2$

So, the magnitude of centripetal acceleration  of the car will be approximately $8\ m/s^2$. Hence, this is the required solution.

Learn more :

Topic : Circular motion.

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