A race car enters the circular portion of a track that has a radius of 63 m. when the car enters the curve at point p, it is tra

Question

1 year ago

14

velling with a speed of 120 km/h that is increasing at 5 m/s2.

1 answer:

frutty [35] · Answer 1 · 2020-09-28T06:05:25+03:00

When car enters a circular path its speed is increasing at rate 5 m/s^2

so this is the tangential acceleration of car

$a_t = 5 m/s^2$

now at the same time since car is moving along the curved path so it must have centripetal acceleration at the same time

It is given by equation

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

$v = 120 km/h = 33.33 m/s$

$R = 63 m$

$a_c = \frac{33.33^2}{63}$

$a_c = 17.64 m/s^2$

so now the net acceleration of the car is given as

$a = \sqrt{a_c^2 + a_t^2}$

$a = \sqrt{5^2 + 17.64^2}$

$a = 18.33 m/s^2$