Tiêu đề Essential Mathematics (Passage based).pdf ✅

PHYSICS Passage # 1 Acceleration of a particle is given by a =v dx dv = dt dv = 2 2 dt d x When acceleration is given by a = f(x) then for find velocity in term of x we write a = v dx dv = f(x) and on integration we get  v v1 v dv =  x x1 f(x) dx But when a is given by a = f(t) then for finding velocity in terms of t we write dt dv = f(t) on integration we get  v v1 dv =  t t1 f(t) dt But when acceleration is given by a = f(v) then we use dt dv = f(v) when we have to find v in terms of t and we used v dx dv = f(v) when we have to find v in terms of x. Q.1 A particle is moving along a straight line with acceleration a = (2t + 1) and at t = 0, v = 0 the velocity of the particle is given by - (A) t2 + t (B) 2t + 1 (C) t (D) 2t2 + t Sol. [A] dt dv = (2t + 1)  v 0 dv =   t 0 (2t 1) dt v = t2 + t Q.2 Acceleration of a particle is given by a = (2x + 1) and at x = 0, v = 0. The speed of particle at x = x is- (A) 2x2 + 1 (B) x2 + 1 (C) 2(x x) 2  (D) None of these Sol. [C] dx dv v = 2x + 1 vdv = (2x + 1) dx  v 0 vdv =   x 0 (2x 1)dx  2 v 2 = x 2 + x Q.3 Acceleration of a particle is given by a = (2v + 1) m/s2 and at t = 0, v = 0. The velocity of particle at t = t is - (A) 2 e –1 2t (B) 2 e 2t (C) e2t –1 (D) v + 1 Sol.[A] dt dv = 2v + 1   v 0 2v 1 dv =  t 0 dt v (v 1) log 2 1 e  = t v v 1 = e2t v + 1 = ve2t 1 = v(e2t – 1) v = e 1 1 2t  Passage # 2 A particle is moving along a straight line and it position is given by x = t3 – t 2 + 1, velocity is given by dt dx and acceleration is given by 2 2 dt d x Q.4 Velocity of the particle at t = 2s is - (A) 2 m/s (B) 4 m/s (C) 6 m/s (D) 8 m/s Sol. [D] x = t3 – t 2 + 1 dt dx = 3t2 – 2t t = 2 s       dt dx = 12 – 4 = 8 m/s Q.5 Acceleration of the particle at t = 1s (A) 2 m/s2 (B) 4 m/s2 (C) 6 m/s2 (D) zero Sol.[B] a = 6t – 2 t = 1 a = 4 m/s2 Q.6 Time at which acceleration is zero - (A) t = 3 4 sec (B) t = 3 2 sec (C) t = 1 sec (D) t = 3 1 sec Sol.[D] 6t – 2 = 0  t = 3 1 s Passage # 3