Answer to Question #93893 in Mechanics | Relativity for kushal

Question

An object starts from rest an accelerates at a constant rate in the +x direction and at some magnitude of velocity, in the +x direction, it has covered a displacement of 6 meters in the +x direction in a time of 3.6 seconds. A different object starts from rest and accelerates at a constant rate in the -x direction with a magnitude of acceleration equal to a factor 3 times less than the first object. When the second object reaches a magnitude of velocity which is a factor of 3 larger than the first objects, was is the magnitude of the second object's?

Answer 1

The formula of velocity from a state of rest with acceleration 'a'

$$u=at \\Under\ the\ terms\ \frac{u_2}{u_1}=3 \ and\ \ \frac{a_1}{a_2}=3 => \frac{t_2}{t_1}=9;$$

By the formula of distance from rest

$$s=\frac{at^2}{2} \\But \ \frac{a_1}{a_2}=3 \ and\ \frac{t_2}{t_1}=9 =>\frac{s_2}{s_1}=12 \\s_2=12s_1=43.2 \ m$$