Answer to Question #163225 in Mechanics | Relativity for M

Question

An object moving with constant acceleration has velocity v = 30.8 m/s when it is at x = 3.6 m and v = 55.1 m/s when it is at x = 6.4 m. What is its acceleration? 

Answer 1

$x_0 = 3.6;V_0=30.8$

$x_1=6.4;V_1=55.1$

$a= \frac{V_1-V_0}{t};t= \frac{V_1-V_0}{a}$

$S_x= x_1-x_0=6.4-3.6=2.8$

$S_x=V_0t+\frac{at^2}{2};$

$S_x=V_0*\frac{V_1-V_0}{a}+\frac{a\frac{(V_1-V_0)^2}{a^2}}{2};$

$S_x=\frac{V^2_1-V^2_0}{2a}$

$a=\frac{V^2_1-V^2_0}{2S_x}=\frac{55.1^2-30.8^2}{2*2.8}=372.2$

Answer: 372.2 m/s²