Emery Hill

2023-02-28

A race car starts from rest on a circular track of radius 400m. The car's speed increases at the constant rate of $0.500m/{s}^{2}$. At the point where the magnitudes of the centripetal and tangential accelerations are equal, determine the speed of the car ?

Valentino Sloan

Beginner2023-03-01Added 3 answers

Centripetal acceleration $a}_{c$ the following is stated

$a}_{c}=R{\omega}^{2}=\frac{{v}^{2}}{R$

where $v$ is linear velocity of the object, $\omega$ is its angular velocity and $R$ is the radius of the circle in which object moves.

Tangential acceleration $a}_{t$ is given to be $=0.500m{s}^{-2}$

Comparing the size of two and putting in a value $R$ we obtain

$\frac{{v}^{2}}{400}=0.500$

$\Rightarrow v=\sqrt{0.500\times 400}=\sqrt{200}=14.142m{s}^{-1}$

$a}_{c}=R{\omega}^{2}=\frac{{v}^{2}}{R$

where $v$ is linear velocity of the object, $\omega$ is its angular velocity and $R$ is the radius of the circle in which object moves.

Tangential acceleration $a}_{t$ is given to be $=0.500m{s}^{-2}$

Comparing the size of two and putting in a value $R$ we obtain

$\frac{{v}^{2}}{400}=0.500$

$\Rightarrow v=\sqrt{0.500\times 400}=\sqrt{200}=14.142m{s}^{-1}$