A car travelling at a constant speed of 30 m/s passes a police car hidden, one second after the speeding car passes the policeman the policeman sets of in pursuit at a constant acceleration of 3m/s. how long does it take the police car to catch the speed driver?
Note by the time the police set off in motion the speeding car has already travelled 30m
Note by the time the police set off in motion the speeding car has already travelled 30m
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The police car catches the speeding car
when the distances are equal:
Distance for speeder = v * t
Distance for police = 0.5 * a * (t^2)
So : v * t = 0.5 * a * (t^2)
And : t = v / (0.5 * a)
So : t = 30 / 1.5
t = 20 seconds + 1 second delay = 21 seconds
when the distances are equal:
Distance for speeder = v * t
Distance for police = 0.5 * a * (t^2)
So : v * t = 0.5 * a * (t^2)
And : t = v / (0.5 * a)
So : t = 30 / 1.5
t = 20 seconds + 1 second delay = 21 seconds
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Basically what you have here is two functions and you are looking for the point of intersection. The function for the constant speed car will be y=30x+30 and the function for the accelerating car will be y=(1/2)3(x^2). so to find the point of intersection you set them equal to each other.
So 30X+30=(1/2)3(x^2) Then just salve for x....
Note* Both functions came from the same equation... D=VT+(1/2)AT^2 Where D= distance V= initial velocity A= acceleration and T=time... To use this in a graph i used y as D and x as t... because it is a function of time. IN the first function we only used D=VT because the acceleration was 0. However you had to account for the extra 30m. so i moved the function 30 up... as seen here (f(X)+30). The second function was D=(1/2)AT^2 because the Initial velocity was 0.
So 30X+30=(1/2)3(x^2) Then just salve for x....
Note* Both functions came from the same equation... D=VT+(1/2)AT^2 Where D= distance V= initial velocity A= acceleration and T=time... To use this in a graph i used y as D and x as t... because it is a function of time. IN the first function we only used D=VT because the acceleration was 0. However you had to account for the extra 30m. so i moved the function 30 up... as seen here (f(X)+30). The second function was D=(1/2)AT^2 because the Initial velocity was 0.
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Let the distance travelled by speeding car is "x" till the police car catches it. So the distance travelled by police car is"x+30"
For Speeding Car x=30t (where t is the time taken to catch the speed driver)
and For Police Car x+30=1/2*3*t²
or 30t+30=3/2*t²
or 3t²=60t+60
or 3t²-60t-60=0
or 3(t²-20t-20)=0
or t²-20t-20=0
Solve the quadratic equation
For Speeding Car x=30t (where t is the time taken to catch the speed driver)
and For Police Car x+30=1/2*3*t²
or 30t+30=3/2*t²
or 3t²=60t+60
or 3t²-60t-60=0
or 3(t²-20t-20)=0
or t²-20t-20=0
Solve the quadratic equation