Presume its an initial condition (s(5)=0) for motion for the second part, but it would help to see the working youre referring to. So its probably simply saying thats the origin for the t>5 phase.
Its worth noting that the second phase is constant acceleration/suvat motion so there is no need to integrate.
Presume its an initial condition (s(5)=0) for motion for the second part, but it would help to see the working youre referring to. So its probably simply saying thats the origin for the t>5 phase. Its worth noting that the second phase is constant acceleration/suvat motion so there is no need to integrate.
But i am confused why they are substituting 10 seconds in the second part, i thought it travels 5 seconds in the first part and 5 in the next so we need to substitute 5 instead? Thank you!
t is not the time duration, it is the time passed. You can solve the question by introducing limits to the integral for the second interval. these limits would be 5 and 10
t is not the time duration, it is the time passed. You can solve the question by introducing limits to the integral for the second interval. these limits would be 5 and 10
similarly for the first interval the limits are 0 and 5.
Youve really just got to work through the maths/sketch stuff. They break it at 5s as for t<5 its a positive displacement (relative to an origin when t=0) and for t>5 its a negative dispacement (relative to an origin when t=5) as the velocity is positive and then negative. For the second phase, you start at t=5 with u=0 and finish at t=10 with v=-10. a=-2. You could use that info with a suvat for S (negative displacement for the second phase) with T=5 (using uppercase just to try and keep things clear), or integrate velocity from t=5 to 10 with s=0 to S. They do the latter.