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Question

Physics Question on Acceleration

Starting from the origin at time t=0t=0, with initial velocity 5j^ms1,5 \hat{ j } \,ms ^{-1}, a particle moves in the xyx - y plane with a constant acceleration of (10i^+4j^)ms2.(10 \hat{ i }+4 \hat{ j }) ms ^{-2} . At time tt, its coordinates are (20 m,y0m)\left. m , y _{0} m \right). The values of tt and y0y _{0}, are respectively

A

4s4 \,s and 52m52\, m

B

2s2\, s and 24m24\, m

C

2s2\, s and 18m18\, m

D

5s5\, s and 25m25\, m

Answer

2s2\, s and 18m18\, m

Explanation

Solution

Given u=5j^m/s,a=10i^+4j^,\vec{ u }=5 \hat{ j } m / s , \vec{ a }=10 \hat{ i }+4 \hat{ j }, final coordinate (20,y0)\left(20, y_{0}\right) in time tt Sx=4xt+12axt2S_{x}=4_{x} t+\frac{1}{2} a_{x} t^{2} 200=0+12×10×t220-0=0+\frac{1}{2} \times 10 \times t^{2} t=2sect=2 sec Sy=uy×t+12ayt2S_{y}=u_{y} \times t+\frac{1}{2} a_{y} t^{2} y0=5×2+124×22=18my _{0}=5 \times 2+\frac{1}{2} 4 \times 2^{2}=18 m 2sec2 \sec and 18m18 m