Question

A car moves in a cirde at the constent speed of 50 m/s and completes one revolution
in 40s, Determine the magnitude of the acceleration of the car.​

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\red{\boxed{\sf Centripetal\: Acceleration (a_c)=7.85m/s^2 }}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Speed (v) = 50m/s

• Time (t) = 40s

$\large\underline\pink{\sf To\:Find: }$

• Magnitude of accleration of the car $\sf{(a_c)=?}$

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We know that ,

${\boxed{\sf Distance (D)=Velocity (v)×Time(t) }}$

$\large\implies{\sf s= 50×40 }$

$\large\implies{\sf s=2000m}$

D = Circumference of the Circle .

$\large{\boxed{\sf D=2πr }}$

$\large\implies{\sf 2000=2×3.14×r}$

$\large\implies{\sf r=\frac{2000}{2×3.14}}$

$\large\implies{\sf r=318.47\:m}$

Now We have to Calculate magnitude of acceleration of the car.

Here the motion is circular .

Therefore Centripetal accleration is present $\sf{(a_c)}$

$\large{\boxed{\sf a_c=\frac{v^2}{r}}}$

$\large\implies{\sf a_c=\frac{(50)^2}{318.47}}$

$\large\implies{\sf a_c=7.85m/s^2}$

$\red{\boxed{\sf Centripetal\: Acceleration (a_c)=7.85m/s^2 }}$

Hence ,

Magnitude of the accleration of the car is $\sf{7.85m/s^2}$

Answer 2

Answer:

We know that, distance = speed*time

So, d = 50*40 = 2000m

d = 2*3.14*r (circumference of the circle)

=> 2000=2*3.14*r

On calculating, r = 318.47 m

In this case, required acceleration will be centripetal acceleration.

So, a = (v*v)/r

        a = (50*50)/318.47

        a = 7.85m/s^2