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 

Guest Jun 22, 2017

1+0 Answers


\(\text{We know that the velocity doubles every interval }\Delta t\text{, hence the velocity}\\ \text{must adhere to the general formula: }v(t)=v_0\cdot2^{t/\Delta t}\text{, with }v_0\text{ the initial}\\ \text{velocity. The distance travelled is of course the integral of the velocity}\\\text{over the time spent travelling:}\\ \Delta x=\int^{t_e}_0v(t)\text{d}t=v_0\int^{t_e}_02^{t/\Delta t}\text{d} t=\frac{v_o\Delta t}{\ln 2}(2^{t_e/\Delta t}-1).\\ \text{We require the time it takes to travel this distance }(t_e)\text{ so we rewrite the}\\ \text{formula to extract }t_e:\\ t_e=\frac{\Delta t}{\ln 2}\ln\left(\frac{\Delta x\ln 2}{v_0\Delta t}+1\right).\\ \text{In order to test our answer we try }\Delta x=0\text{ this should give us }t_e=0,\\ \text{and indeed it does.}\)

Honga  Jun 23, 2017

16 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details