Consider your lunch time trip again(figure 2.1) For how many seconds did you acc

Question

Consider your lunch time trip again(figure 2.1) For how many seconds did you accelerate at 2.0 m/s^2 to get to a cruising speed of 15 m/s? You will have traveled 56 m toward the gas station. When you brake to pull into the gas station you do so at -2.0 m/s^2. How much time does that take? The distance from the University to the gas station that you traveled at constant speed is one mile minus twice the accelerating distance(1.00 mile -2 times 56 m). How much time does it take this distance at the constant cruising speed of 15 m/s? What fraction of the total time in going from the university to the gas station were you accelerating? This illustrates that we spend very little time in our lives accelerating and mostly travel at constant speed.

Explanation / Answer

1 mile = 8/5 km = (8/5)*1000 = 1600 m

(a)

initial speed vi =0

final speed vf = 15 m/s

vf = vi + a*t

15 = 0 + 2*t

t = 7.5 s

(b)

vf = 0

vi = 15 m/s

a = -2

vf = vi + at

0 = 15 - 2*t

t = 7.5 s

(c)

here acceleration a = 0

x = v*t

1600 - (2*56) = 15*t

t = 99.2 s

(d)

fraction = (7.5+7.5)/(7.5+99.2+7.5) = 0.13

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