Question

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} )

# (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']

Solution