(D) 1:3:9 x. next 10 s is x, and th...
Question
Fullscreen

(D) 1:3:9 x. next 10 s is x, and the last 10 s is x,. Then x, X,. 1, (C) 1:3:5 (A) 1:2:4 (B)1:2:5 A particle travels 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration what is the distance travelled in next 2 sec. (D) None of above (A) 8.3 m (B) 9.3 m (C) 10.3 m A particle is thrown upwards from ground. It experiences a constant resistance force which can to the time of descent is (g=10 m/s']

JEE/Engineering Exams
Physics
Solution
103
Rating
4.0 (1 ratings)
Fullscreen
withitial welority of pactur be ( =u ) for fust 5 sec motion ( s_{5}=10 mathrm{m} ) ( S=u t+frac{1}{2} a t^{2} ) ( Rightarrow quad 0=5 u+frac{1}{2} a(5)^{2} ) ( Rightarrow 2 u+5 a=4 quad-0 ) For fist 8 sec of motion ( S_{B}=20 mathrm{m} ) ( 20=8 u+frac{1}{2} a(8)^{2} ) ( Rightarrow 2 u+8 a=5-(2) ) By solung O and (2), we get ( u=frac{7}{6} m / s ) and ( a=frac{1}{3} m / s^{2} ) Now distanc tranelled sy pacticle in 10 sec total ( s_{10}=u times 10+frac{1}{2} a(10)^{2} ) By substituting tul value of a anda in sio , ure get, ( s_{10}=28 cdot 3 mathrm{m} ) i. Distance in last 2 ser in ( begin{aligned}=S_{10}-S_{8} &=28.3-20 &=8.3 mathrm{m} &=text { option } A) end{aligned} )
Quick and Stepwise Solutions Just click and Send Download App OVER 20 LAKH QUESTIONS ANSWERED Download App for Free