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Question

Physics Question on Motion in a straight line

A body covers a distance of 4m4\, m in 3rd3^{rd} second and 12m12 \,m in 5th5^{th} second. If the motion is uniformly accelerated, how far will it travel in the next 33 seconds ?

A

10m10\,m

B

30m30\,m

C

40m40\,m

D

60m60\,m

Answer

60m60\,m

Explanation

Solution

S3=u+a2(2×31)=4S_{3}=u+\frac{a}{2}\left(2\times3-1\right)=4 or u+52a=4u+\frac{5}{2} a=4 S5=u+a2(2×51)=12S_{5}=u+\frac{a}{2}\left(2\times5-1\right)=12 or u+92a=12u+\frac{9}{2}a=12 On solving, u=6ms1,a=4ms2u=-6\,m\,s^{-1}, a=4\,m\,s^{-2} Distance travelled in next 33 seconds =S8S5=S_{8}-S_{5} =[6×8+12×4×(8)2][6×5+12×4×(5)2]=\left[-6\times8+\frac{1}{2}\times4\times\left(8\right)^{2}\right]-\left[-6\times5+\frac{1}{2}\times4\times\left(5\right)^{2}\right] =8020=60cm=80-20=60\,cm