Question

Plz solve it fast with full process. I will mark u brainiest & give u thanks​

Answer 1

$\huge \tt \red{\underline{Solution:}}$

Net acceleration is due to braking and centripetal acceleration

Due to Braking,

$a_{T} = 3 m/ {s}^{2}$

Speed of the cyclist,

$v = 20m/s$

Radius of the circular turn,

$r = 100m$

Centripetal acceleration (acceleration on circular turn) is given as:

$a_{c} = \frac{ {v}^{2} }{r} \\ = \frac{ {20}^{2} }{100} \\ = \frac{ \cancel{400}}{ \cancel{100}} \\ = 4m/ {s}^{2}$

answer : 4 m/s^2

The resultant acceleration a is given by:

$a = \sqrt{( {a_{c}}^{2} + {a_{t}}^{2})} \\ = \sqrt{( {4}^{2} + {3}^{2})} \\ = \sqrt{(16 + 9)} \\ = \sqrt{25} \\ = 5m/{s}^{2}$