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.
Answers
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:
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