Suppose that we have an object that moves into a Δx path. Its speed doubles as a period of Δt passes. No matter the distance of the path, find a way to express the time that it will take the object to go from xstart to xfinish with Δx and Δt being variables
+169 \(\text{We know that the velocity doubles every interval }\Delta t\text{, hence the velocity}\\ \text{must adhere to the general formula: }v(t)=v_0\cdot2^{t/\Delta t}\text{, with }v_0\text{ the initial}\\ \text{velocity. The distance travelled is of course the integral of the velocity}\\\text{over the time spent travelling:}\\ \Delta x=\int^{t_e}_0v(t)\text{d}t=v_0\int^{t_e}_02^{t/\Delta t}\text{d} t=\frac{v_o\Delta t}{\ln 2}(2^{t_e/\Delta t}-1).\\ \text{We require the time it takes to travel this distance }(t_e)\text{ so we rewrite the}\\ \text{formula to extract }t_e:\\ t_e=\frac{\Delta t}{\ln 2}\ln\left(\frac{\Delta x\ln 2}{v_0\Delta t}+1\right).\\ \text{In order to test our answer we try }\Delta x=0\text{ this should give us }t_e=0,\\ \text{and indeed it does.}\)
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