Answer to Question #167009 in Mechanics | Relativity for lisa

Question

A motorcycle slows down on a sharp horizontal turn, going from 30.0 m/s to 10.0 m/s in 15.0 s . The radius of the curve is 166 m. Assuming the motorcycle slowing down during this time at the constant rate, find the magnitude of the total acceleration in m/s²Â at the moment the motorcycle speed reaches 10.0 m/s. 

Answer 1

$u=30m/s$

$v=10m/s$

$t=15 s$

$r=166m$

$\Rightarrow v=u+a_tt$

$\Rightarrow a_t=-\dfrac{4}{3}m/s^2$

Centripetal acceleration, $a_c=\dfrac{v^2}{r}$

$\Rightarrow a_c=\dfrac{(10)^2}{166}$

$\Rightarrow a_c=0.60m/s^2$

Total Accceleration, $a_{total}=\sqrt{a_c^2+a_t^2}$

$\Rightarrow a_{total}=\sqrt{(0.60)^2+(1.33)^2}$

$\Rightarrow a_{total}=1.45m/s^2$