S2 31

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A motorcycle cop, parked at the side of a highway reading a magazine, is passed by a woman in a red Ferrari 30?

A motorcycle cop, parked at the side of a highway reading a magazine, is passed by a woman in a red Ferrari 308 GTS doing 90.0 km/h. After a few attempts to get his cycle started, the officer roars off 2.00 s later. At what average rate must he accelerate if 114 km/h is his top speed and he is to catch her just at the state line 2.30 km away?

__ m/s^2

Speed of the frerai=90km/h= 90000/3600= 25m/s
Total Distance in the system: 2.3km=2300m
So time taken for the Ferari to reach 2300m=2300/25=92s

The max. speed of the Police byke V =114Km/hr.= 31.67m/s
The police has taken 2s during kick starting.

So total time taken by byke= 92-2 => T= 90s

Theoritically the the police should go with a constant speed of 2300/90= 25.55m/s to catch the ferari.

But Practically, Byke does not acquire the 25.55m/s of speed when started.
So there is an accelaration from the Starting of Byke upto time say 'n' and 'n' th second the byke is considered to get the top speed=114km/h=31.67m/s as the police can not ride it above top speed.

and it will run at a constant speed on 31.67m/s until it catches the Ferari.
i.e, It will run for T-n= 90-n second at constant speed 31.67m/s

Here for Byke: Initial speed U=0, Final speed V=31.67m/s
acceleration=a and time ='n'

V=U+at=0+an=an => a= V/n ---------------eq.1

S= S1+S2= V^2/2a + V*(90-n)= 0.5Vn+90V-nV=- 0.5V n +90V

2300= 0.5Vn+90V

2300= 0.5*31.67*n +90*31.67= -15.835n+2850.3
=>2300=2850.3-15.835n=>550.3=15.835n
=>n=34.8 s

So a=V/n =31.67/34.8=0.91m/s^2
So desired accelaration= 0.91m/s^2 --------Answer

Check Answer:
S1= 1/2 an^2= 0.5*0.91*34.8*34.8=551.06m
S2=31.67*(90-34.8)=1748.184

S1+S2=2299.44mtrs which is equal to 2.3km

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