3 Answers | Add Yours
Note that since the distance from the origin increases from 10 up to 65 metres (for the cyclist) or 65 km (for the train) and then decreases back to 0, the cyclist would have to turn round and go back down the hill (or continue over a symmetric hill) and continue from his start point (or symmetric start point) for 10 metres toward the origin (or symmetric origin) of travel. The train could have been 10km from the original station (origin) initially and then, after stopping at the next station, returned back on the same track all the way to the original station.
The object must move 55 units in a particular direction/route and then return on the same route (the same in a vertical direction or horizontal direction, or both) and continue on for 10 units distance.
If the units are km/minute and minutes this could be a ``high-speed bullet train (possibly of the future!).
The train pulls out of a station and after 23 mins reaches 4km/min = 240km/hr. It then decelerates for 6 mins and stops at a station. Passenger alight and board for 3 mins. The train then pulls out of the station and accelerates for 5 minutes up to 8.5km/min (approximately 500km/hr). It then decelerates for 10mins to the next station.
To make the graphs more realistic make them more curved as sudden changes in speed aren't common in reality. In this case you could add S-curves to indicate acceleration and deceleration:
2a) distance-time graph
If the units are metres per second and time in seconds then this could be a cyclist. They accelerate for 22s (max speed 4m/s) and then, encountering a hill decelerate for 5s. They stop for 3s to admire the view :) and then accelerate downhill for 5s without breaking (max speed 8.5m/s). It takes them 10s to come to a complete halt.
We’ve answered 318,955 questions. We can answer yours, too.Ask a question