A car starts from rest and travels for 5.0 seconds with uniform acceleration of 1.5 m/s^2. The driver then…?

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applies the brakes, causing a uniform acceleration of 2.0 m/s^2 in the negative direction. If the brakes are applied for 3.0 seconds, how fast is the car going at the end of the braking period, and how far has it gone?I did this problem for homework, I just want to make sure my answer and math is right

Madie Hilpert · Accepted Answer

First lets compute the distance from rest and before applying the brakes and the velocity at the instant of applying the brakes: For distance:d=Vt+1/2at^2where V=starting velocityd=(0)(5)+1/2(1.5)(5)^2d=18.75 meter For the velocity at the instant of applying the brakes:v=V+atv=0+(1.5)(5)v=7.5 m/sThen the driver applies the brakes for deceleration of 2.0 m/s^2 for 3 secondsthe final velocity will beV(final)=v+AT where A= deceleration of 2.0 m/s^2 and T= time of braking of 3 sV(final)=7.5+(-2)(3) ---> A is negative because it is deceleration or acceleration at negative directionthereforeV(final)=1.5 m/sThe distance from the instant of braking to the final position will be:D=vT+1/2AT^2D=(7.5)(3)+1/2(-2)(3)^2D=22.5+1/2(-2)(9)D=22.5-9D=13.5 mTherefore the total distance istotal distance=d+Dtotal distance=18.75+13.5total distance=32.25 meters

Madie Hilpert · Answer

effortless in simple terms use the acceleration equation (vf = vi +at) for the first 7 seconds then use it for the braking subtract both solutions and also you get section a. for b in simple terms use the acceleration equation that includes a distance ingredient to it. (d = vit + a million/2at^2)

Madie Hilpert · Answer

U=0t=5seca=1.5m/s^2V=U+at =0+1.5*5=7.5m/sDitance S=Ut +1/2*at^2 =0+1/2 *1.5*5^2 18.75mNow retardationu=7.5m/s (last vel in acceleration)a=-2m/s^2t=3secv=u-at7.5-2*31.5m/s (Last velocity)s=ut-1/2at^2=7.5*3-1/2*2*3^2= 22.5-9=13.5m(Last velocity)= 1.5m/sTotal distance=18.75+13.5 =32.25m

Madie Hilpert · Answer

10.5 m/s for final velocity78 m distance

Kelsie Sawayn

A car starts from rest and travels for 5.0 seconds with uniform acceleration of 1.5 m/s^2. The driver then…?

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