A position of particle is `x=`at`+bt^(2)-ct^(3)` find out velocity when acceleration is zero (1) `v=a+b^2/(3c)` (2) `v=a-b^2/(3c)` (3) `v=2a-b^2/(3c)` (4) None of these

If (a)/(2)=(b)/(3)=(c)/(4), then a:b:c=2:3:4(b)4:3:23:2:4(d) None of these

The position of a particle as a function of time t, is given by x(t)=at+bt^(2)-ct^(3) where a,b and c are constants. When the particle attains zero acceleration, then its velocity will be :

If a/2=b/3=c/4, then a : b : c= (a) 2:3:4 (b) 4:3:2 (c) 3:2:4 (d) None of these

Displacement (x) and time (t) of a particle in motion are related as x = at + bt^(2) -ct^(3) where a,b, and c, are constants. Velocity of the particle when its acceleration becomes zero is

If the velocity of a particle is v = At+Bt^(2) , where A and B are constant then the distance travelled by it between 1 s and 2 s is - (A) (3A +7B) (B) (3)/(2)A+(7)/(3)B (C) (A)/(2)+(B)/(3) (D) (3)/(2)A+4B

The velocity time relation of a particle is given by v = (3t^(2) -2t-1) m//s Calculate the position and acceleration of the particle when velocity of the particle is zero . Given the initial position of the particle is 5m .

The velocity time relation of a particle is given by v = (3t^(2) -2t-1) m//s Calculate the position and acceleration of the particle when velocity of the particle is zero . Given the initial position of the particle is 5m .

In the expansion of (x-(1)/(3x^(2)))^(9), the term independent of x is T_(3) b.T_(4) c.T_(5) d.none of these

If v=3t^(2) - 2t +5 , where v is velocity (in m/s) and t is time (in second). Find the displacement (in m) of particle in third second. (A) 19 (B) 24 (C) 9 (D) None of these

If the two equations ax^(2)+bx+c=0 and 2x^(2)-3x+4=0 have a common root then (A) 6a=4b=-3c(B)3a=-4b=3c(C)6a=-4b=3c (D) none of these