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a particle travels 10 m in first 5 sec and 10 m in next 3 seconds . assuming constant acceleration what is the distance travelled in next 2 seconds. |
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Answer» Answer: we know that initial velocity of a particle will be 0. 1)GIVEN time =5 seconds, Distance =10 m/s s=ut+1/2a*t^2 10=u×5+1/2*a*(5)^2 10=5u+25a/2 10-(5u+25a/2)=0 -5(2u+5a-4)=0 2u+5a-4=0 2u +5a=4.......(1) (2) given time =5+3=8 seconds, Distance =20m s=ut+1/2a*t^2 20=8u+1/2*a(8)^2 20=8u+64a/2 20-(8u+64a/2)=0 20-8u-32a=0 -4(2u+8a-5)=0 2u+8a-5=0 2u+8a=5........(2) on solving (1) and(2) we get 4-5=2u+5a-2u-8a -1=-3a a=1/3.....(3) substitute (3) in (2) we get 2u+8a=5 2u+8(1/3)=5 2u+8/3=5 6u+8=15 6u=7 u=7/6.........(4) therefore the distance covered in first 10 seconds s=ut+1/2at^2 =7/6*10+1/2*1/3(10)^2 =70/6+100/6 =170/6 =28.3m therefore the distance TRAVELED by the particle in next 2 seconds =28.3-20 =8.3m Explanation: please do follow me |
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