Question

(D) SS OTO 5. ody has an initial velocity of 3 ms-1 and has a constant acceleration of 1 ms-2 normal to the direction of the initial velocity. Then its velocity, 4 second after the start is (A) 7 ms- along the direction of initial velocity (B) 7 ms- along the normal to the direction of the initial velocity (C) 7 ms-1 mid-way between the two directions (D) 5 ms- at an angle of tan with the direction of the initial velocity

JEE/Engineering Exams

Physics

Solution

114

4.0 (1 ratings)

we will solve this problem in two direction on in is horizontal and one in vertical..... in horizontal velocity will unchanged after 4 sec i.e ( mathrm{Vx}=3 mathrm{m} / mathrm{s} ) in vertical motion intial velocity is zero and so final velocity in vertical motion ( mathrm{Vv}=mathrm{uy}+mathrm{at} ) ( mathrm{Vy}=0+1^{star} 4=4 mathrm{m} / mathrm{s} ) since acceleration is normal to the initial velocity so net velocity after 4 sec ( =v^{prime}= ) ( left(4^{wedge} 2+3^{wedge} 2right)^{wedge} 1 / 2=5 m / s ) direction theta ( = ) tan inverse ( (mathrm{vy} / mathrm{vx})= ) tan inverse ( (4 / 3) ) Thanks & Regards, Nirmal singh Askiitians faculty