Question

A cyclist increases her speed uniformly from 3m/s over a distance of 40m. Calculated the magnitude of the acceleration of the cyclist over this distance

Answer 1

Condition:

$v_{0}=0 \frac ms$  

$v=3 \frac ms$

$l=40 m$

$a- ?$

Solution:

For uniformly accelerated motion:

$l=\frac{(v^2-v_0^2)}{2a}$

but $v_{0}=0 \frac ms$ , then

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

where

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

Substitute the numbers.

$a=\frac{3^2}{2\times40}=\frac {9}{80}=0.1125\frac {m}{s^2}$

Answer: $0.1125 \frac{m}{s^2}$ .