How Do You Calculate Distance Traveled and Minimum Stopping Acceleration?

[Girldistracted]

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??

 

[himanshu121]

For first part you have the graph why don't you find the area under the curve v-t from t=1.8 to t= 3.6

 

[M98Ranger]

For the first question, you can use the equation for average velocity: v = (vf + vi)/2, where vf is the final velocity and vi is the initial velocity. Rearrange the equation to solve for distance traveled: d = v * t. Plug in the values for v and t, and you will get the distance traveled in that time interval.

For the second question, you can use the equation for distance: d = vi * t + 1/2 * a * t^2, where vi is the initial velocity, t is the time, and a is the acceleration. In this case, vi is 28.7 m/s, t is the time it takes for the car to stop (unknown), and d is 107 m. Rearrange the equation to solve for a, and you will get the minimum acceleration needed for the car to stop without hitting the deer.