- #1
Girldistracted
- 1
- 0
The Graph above shows the speed of a car traveling in a straight line as a function of time. (Sorry no graph) The car accelerates uniformly and reaches a speed Vb of 3.20 m/s in 8.00 seconds. Calculate the distance traveled by the car from a time of 1.80 to 3.60 seconds.
Im not sure what equations to use to find this...i thought i would find the change in time and then multiply that by the final velocity but that was wrong and i don't know what else to use bc the equations i have require another velocity or acceleration.
My second question is A driver in a car traveling at a speed of 64.2 mi/h sees a deer 107 m away on the road. Calculate the minimum constant acceleration that is necessary for the car to stop without hitting the deer (assuming that the deer does not move in the meantime).
I changed the speed to 28.7m/s. I used the equation
v^2=v0^2+2a(x-x0) and solved for a but that was wrong...can anyone help??
Im not sure what equations to use to find this...i thought i would find the change in time and then multiply that by the final velocity but that was wrong and i don't know what else to use bc the equations i have require another velocity or acceleration.
My second question is A driver in a car traveling at a speed of 64.2 mi/h sees a deer 107 m away on the road. Calculate the minimum constant acceleration that is necessary for the car to stop without hitting the deer (assuming that the deer does not move in the meantime).
I changed the speed to 28.7m/s. I used the equation
v^2=v0^2+2a(x-x0) and solved for a but that was wrong...can anyone help??