Question

A new car is tested on a $200-m$-diameter track. If the car speeds up at a steady $1.5m/s^2$ , how long after starting is the magnitude of its centripetal acceleration equal to the tangential acceleration?

Short answer

The magnitude of its centripetal acceleration equal to the tangential acceleration after$8.2s$.

Step-by-step solution

Given Information

Diameter of the track : $d=200\mathrm{m}$.

Acceleration of the car $a_t=1.5\mathrm{m}/{\mathrm{s}}^2$.

Explanation

According to the information:

$d-$The diameter of the truck $=200m$

$a_t-$Tangential acceleration$=1.5m/s^2$

The centripetal acceleration$a_c$ is given as follow:

$a_c=\frac{v^2}{r}$

Where,

$v-$The velocity of the car

$r-$The radius of the track

$t-$The time

Applying the equation to find time

Centripetal acceleration $a_c$$=$Tangential acceleration $a_t$

$a_t=\frac{v^2}{r}$

$v=\sqrt{a_t\times r}$ (1)

The initial velocity of the car is zero.

Therefore, the tangential acceleration is given by kinematic equation as follow:

$v=a_t\times t$ (2)

From equation 1 and 2, we get

$\sqrt{a_t\times r}=a_t\times t$

$t=\sqrt{\frac{r}{a_t}}$

$t=\sqrt{\frac{d}{2a_t}}$

$t=\sqrt{\frac{200}{2\times 1.5}}$

$t=\sqrt{66.67}$

$t=8.2s$

Final Answer

The magnitude of its centripetal acceleration equal to the tangential acceleration after$8.2s$.