Asked by Jade
A motorist with an expired license tag travels at a constant speed of 17.2 m/s down a street, and a policeman on a motorcycle, taking another 4.08 s to finish his donut, gives chase at an acceleration of 2.71 m/s2.
Answers
Answered by
Damon
x is distance from donut shop
t is time police chases, so car is 4.08 seconds more
when they are both x meters from shop:
x = 17.2 ( t+4.08)
x = (1/2) 2.71 t^2
when they meet
1.35 t^2 = 17.2 t + 70.2
or
1.35 t^2 -17.2 t - 70.2 = 0 solve quadratic
https://www.mathsisfun.com/quadratic-equation-solver.html
so t = -1.78 or t = 3.16
negative time was before the car started so use 3.16 seconds
then the car went 3.16 + 4.08 = 7.24 seconds
and they meet 17.2 * 7.24 = 125 meters from donuts
t is time police chases, so car is 4.08 seconds more
when they are both x meters from shop:
x = 17.2 ( t+4.08)
x = (1/2) 2.71 t^2
when they meet
1.35 t^2 = 17.2 t + 70.2
or
1.35 t^2 -17.2 t - 70.2 = 0 solve quadratic
https://www.mathsisfun.com/quadratic-equation-solver.html
so t = -1.78 or t = 3.16
negative time was before the car started so use 3.16 seconds
then the car went 3.16 + 4.08 = 7.24 seconds
and they meet 17.2 * 7.24 = 125 meters from donuts
Answered by
henry2,
d1 = d2.
17.2*4.08 + 17.2*T = 0.5*a*T^2.
70.2 + 17.2T = 0.5*2.71*T^2,
1.35T^2 - 17.2T - 70,2 = 0,
T^2 - 12.7T - 52 = 0, Use Quad. Formula.
T = (12.7 +- sqrt(162 + 208))/2 = (12.7 +- 370)/2 = 16, and -3.3 s.
T = 16 s.
d = 0.5*2.71*16^2 = 347 m.
17.2*4.08 + 17.2*T = 0.5*a*T^2.
70.2 + 17.2T = 0.5*2.71*T^2,
1.35T^2 - 17.2T - 70,2 = 0,
T^2 - 12.7T - 52 = 0, Use Quad. Formula.
T = (12.7 +- sqrt(162 + 208))/2 = (12.7 +- 370)/2 = 16, and -3.3 s.
T = 16 s.
d = 0.5*2.71*16^2 = 347 m.
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