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Grade 12Mechanics

A police officer sits on a parked motorcycle. A car travelling at a constant speed of v_o= 40.0 m/s passes by at t=0. At that instant the officer accelerates the motorcycle at a constant rate and at time t₁ = 20.0s overtakes the speeder.
(a) Find the acceleration of the motorcycle.
(b) Find the speed of the motorcycle at that instant it overtakes the car.

himani

8 Years agoGrade 12

3 Answers

vishal

8 Years ago

after 20 s the car has travelled 20 x 40=800 mso the police has to travel 800 m to catch that carinitial velocity of car is 0 displ,s=0+1/2(a)t2where t=20 sfind out sv2=u2+2as=0+2(

Venkat

8 Years ago

For speeder ->v_{0}^{s}=40 m/s = v^{^{s}} , a^{s}=0 \\For Policeman->v_{0}^{p}=0 \\ \\ (a)x^{s}=v^{s}t=40(20)=800m \\x^{p}=v_{0}^{p}+0.5a^{p}t^{2}->0+0.5a^{p}(20)^{2}->200a^{p} \\when policeman catches the speederx^{p}=x^{s} \\800=200a^{p}->a^{p}=4m/s^{2} \\ \\(b) v^{p}=a^{p}t=4(20)=80 m/s

Venkat

8 Years ago

The answer is below.

Please ignore my above answer as there was issue with the latex solution. The solution is same but, just put in latex to look ok.

$For speeder ->v_{0}^{s}=40 m/s = v^{^{s}} , a^{s}=0 \\For Policeman->v_{0}^{p}=0 \\ \\ (a)x^{s}=v^{s}t=40(20)=800m \\x^{p}=v_{0}^{p}+0.5a^{p}t^{2}->0+0.5a^{p}(20)^{2}->200a^{p} \\when policeman catches the speederx^{p}=x^{s} \\800=200a^{p}->a^{p}=4m/s^{2} \\ \\(b) v^{p}=a^{p}t=4(20)=80 m/s$

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