Question

A car moves at a constant acceleration and crosses a path of 400m. What is the acceleration if the car takes 10 seconds to cross the third 100m path?

Answer 1

$\text{let for point } P \text{ middle of the 400m path}$

$x_0= 0;v_0>0$

$s(t) = v_0t +\frac{at^2}{2}$

$s(10)=100$

$s(10)=10v_0+50a$

$v_0+5a-10 = 0\ (1)$

$s_1=\frac{v_1^2-v_0^2}{2a}$

$\text{for }s_1=200(\text{ the car has reached the end of the 400m path}):$

$v_1 = 0$

$200=\frac{-v_0^2}{2a}$

$a= -\frac{v_0^2}{400}$

$\text{Substitute a into expression (1)}$

$v_0-\frac{v_0^2}{80}-10 = 0$

$v_0^2-80v_0+800 = 0$

$v_0= 11.72\frac{m}{s}; a = -\frac{v_0^2}{400}= -0.34\frac{m}{s^2}$

$v_0'= 68.28\frac{m}{s}; a = -\frac{v_0^2}{400}= -11.6\frac{m}{s^2}$

$\text{Answer: }-0.34\frac{m}{s^2} \ or\ -11.6\frac{m}{s^2}$