My question pertains to the stimulator on question 4 found on this website. http

Question

My question pertains to the stimulator on question 4 found on this website.

http://media.pearsoncmg.com/bc/aw_young_physics_11/pt1a/Media/DescribingMotion/AnalyMotUsingGraphs/Main.html

Question 4: Stopping Time and Stopping Distance

Open and run the simulation of the motion graph. Try the following slider settings:

xo = -15m, vo = +16m/s, a = -4m/s2

Predict the time interval needed for the car to stop. Enter your prediction here:

__ ___________________________

Then run the simulation and note on the velocity-versus-time graph, the time when the velocity is zero. Does this agree with your prediction?

Now, calculate the average velocity during the time interval while the car is stopping.

To find the stopping distance, multiply the average velocity by the stopping time. Does this stopping distance match the distance traveled by the car while stopping?

Repeat this problem for the following slider settings:

xo = +15m, vo = -16m/s, a = +4m/s2

xo = -15m, vo = +16m/s, a = -4m/s2

Explanation / Answer

Given xo = -15m, vo = +16m/s, a = -4m/s2

let after time t it will stop

When the car will stop, V = 0

Now we know V = V0 + at   => 0 = 16 -4t => t = 4 sec

During this time interval, the total displacement travelled by the car be

X = V0t + 1/2 at2 = 16*4 - 0.5*4*16 = 32 meter

so average velocity = total displacement/time = 32/4 = 8 m/s

Now stoping distance = average velocity*time = 8*4 = 32 meter which match the distance traveled by the car while stopping

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