A particle is moving with uniform acceleration. Its displacement at an – InterviewSolution

InterviewSolution

1.	A particle is moving with uniform acceleration. Its displacement at any instant t is given by `s=10t+49t^2` .find (i) initial velocity (ii) velocity at `t=3 `second and(iii) the uniform acceleration.
Answer» acceleration is constant `S=10t+49t^2` `V=ds/dt=d/dt(10t+49t^2)` `V=10+98t` 1)Initial velocity `V=10+98t` `V=10`m/s 2)Velocity at t=3sec `V=10+98t` `V=10+98(3)` `V=304`m/s 3)Constant acceleration `a=(dV)/dt` `a=d/dt(10+98t)` `a=98m/s^2`.

Discussion

No Comment Found

Related InterviewSolutions

33 The position vector of a particle moving in x-y plane is given by r (t 4) i (t 4) j. Find (a) Equation of trajectory of the particle Albo in a (b) Time when it crosses x-axis and y-axis
A particle is moving with uniform acceleration. Its displacement at any instant t is given by `s=10t+49t^2` .find (i) initial velocity (ii) velocity at `t=3 `second and(iii) the uniform acceleration.
If `y=(log)_(sinx)(tanx),t h e n(((dy)/(dx)))_(pi/4)"is equal to"`(a)`4/(log2)`(b) `-4log2`(c)`(-4)/(log2)`(d) none of these
Find the derivative of the following functions from first principles: (i) ` x` (ii) `(-x)^(-1)` (iii) `s in (x + 1)` (iv) `cos(x-pi/8)`Find derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s a
`(3x+5)(1+tanx)`
Find the derivative of `sqrt(tanx)` using first principles.
If f(x) = { `sin[x] /[x],[x] != 0 ; 0, [x] = 0}` , Where[.] denotes the greatest integer function, then `lim_(x rarr 0) f(x)` is equal toA. 1B. 0C. `-1`D. Does not exist
`lim_(x->1)[(2x-3)(sqrtx-1)]/[2x^2+x-3]`A. `1/10`B. `-1/10`C. 1D. None of these
`lim_(y to 0) ((x+y)sec(x+y)-xsecx)/(y)`
If `y=(1)/(3x^(3))` then prove that `3y+x(dy)/(dx)=0`.