A car is trailing a truck on a two way, two lane rural roadway. The truck is moving at a constant speed of 70km/h. the car has decided to pass the truck at a starting speed of 68km/h. as the car accelerated and just encroached the right lane it sighted an opposing vehicle moving at a constant speed of 75km/h. at what distance should this opposing vehicle be at this instance of time so that the car can safely execute the passing maneuver? Minimum required distance between the passing car (at the end of maneuver) and the opposing vehicle is 50m. * *The acceleration rate of the car until encroaching on the right lane is 0.53m/s². The time taken for encroachment on the right lane is 5sec. *The distance from front of truck to front of passing car after it encroached to the right lane is 20m. *The required distance between the passing car and the passed truck at the end of the maneuver should be 25m (front of car to front of truck) .Please dear do it
u r way better than any transportation
teacher i hv ever heard
A car is trailing a truck on a two way, two lane rural roadway. The truck is moving at a constant speed of 70km/h. the car has decided to pass the truck at a starting speed of 68km/h. as the car accelerated and just encroached the right lane it sighted an opposing vehicle moving at a constant speed of 75km/h. at what distance should this opposing vehicle be at this instance of time so that the car can safely execute the passing maneuver?
Minimum required distance between the passing car (at the end of maneuver) and the opposing vehicle is 50m.
* *The acceleration rate of the car until encroaching on the right lane is 0.53m/s².
The time taken for encroachment on the right lane is 5sec.
*The distance from front of truck to front of passing car after it encroached to the right lane is 20m.
*The required distance between the passing car and the passed truck at the end of the maneuver should be 25m (front of car to front of truck)
.Please dear do it
11:27 Where he get the 30 from 2g tho?
The 30 is equal to 2g divided by the conversion from fps to mph..... 30 = 2*32.2/((5280/3600)^2)
I believe it is for the conversions.