Chapter 05, I 028 Incorrect A ta, th t eeghs i 2 \" io. ,s naaie moong at steed

Question

Chapter 05, I 028 Incorrect A ta, th t eeghs i 2 " io. ,s naaie moong at steed of 44 bayh when tte brakes are atened and the car brought to·stop 16 m. Assurmang that the force that stops the car magnutude of that force and (b) the tme requred for the change in speed. If the initial speed is doubled, and the car expenences the same force during the braking, by what factors are (c) the stopping datance and (d) the stopping time maltiphed? (There could be a lessen here about the danger of driving at high speeds.)

Explanation / Answer

U^2 = -2aS. where U the initial velocity, a the acceleration and S is the distance travelled. V the final velocity = 0.
U = 44km/h = 44*10 /36 = 12.22 m/s
a = - U^2/ 2S = - 12.22^2 / 2*16 = -4.7 m/s^2.
Force = ma = 1.2e4*(-4.7) = -56400 N
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V = U + at
0= 12.22 + (-4.7) t.
t = - 12.22/ (-4.7) = 2.6 s
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Force remains the same; the acceleration will be the same. S will change in proportion to the square of the initial velocity U. S = U^2 / a Since U is doubled S will be 2*2 = 4 times the previous one.
S1 = 4*16 = 64 m
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Since acceleration is the same,
Time t = initial velocity / acceleration. Since initial velocity is doubled, time will be doubled.
The multiplication factor is 2.

Or since the distance is 4 times the previous one, time will be 4 = 2. time the previous one.

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