Question

A car enters a curved road in the form of a quarter of a circle, the path length being 200 m. Its speed at the entrance is 18 km/h but when it leaves, it increases to 54 km/h. If the car is travelling with constant acceleration along the curve, the acceleration when the car leaves the curved road is:

Answer 1

Answer:

1.84 $m/s^{2}$

Explanation:

acceleration = resultant of tangential acceleration and centripetal acceleration

tangential acceleration =>

u = 18 km/h = 5 m/s

v = 54 km/h = 15 m/s

s = 200 m

Using $v^{2} - u^{2} = 2as$, we get a = 0.5 $m/s^{2}$.

Tangential acceleration = 0.5 $m/s^{2}$

$\frac{2\pi r}{4}$ = 200 m                             [ Quarter of a circle ]

=> r = $\frac{400}{\pi }$ m

Centripetal acceleration =>

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

a = $\frac{15^{2} }{\frac{400}{\pi } }$ $m/s^{2}$

a = 1.76 $m/s^{2}$

Total acceleration = $\sqrt{0.5^{2} +1.76^{2} }$ $m/s^{2}$

                              = 1.84 $m/s^{2}$