A particle is moving with uniform acceleration. Its position (x) is given in term of time (t in sec) as x= '(5t^2 +4t +8)' m, then

A particle is moving with a uniform accerleration. Its displacement at any instant t is given by s=10t +4.9 t^2 . What is the uniform acceleration?

A particle is moving in a straight line. The variation of position ‘x’ as a function of time ‘t’ is given as x = (t^3 – 6t^2 20t 15) m. The velocity of the body when its acceleration becomes zero is :

A particle starts from origin with uniform acceleration. Its displacement after t seconds is given in meter by the relation x=2+5t+7t^2 . Calculate the magnitude of its a. Initial velocity b. Velocity at t=4s c. Uniform acceleration d. Displacement at t=5s

A particle starts from origin with uniform acceleration. Its displacement after t seconds is given in metres by the relation x = 2 + 5 t + 7t^2 Calculate the magnitude of its (i) initial velocity (ii) velocity at t =4 s (iii) uniform acceleration (iv) displacement at t =5 s.

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.

(i) A particle is moving in three dimensions.Its position vector is given by vecr=6hati+(3+4t)hatj-(3+2t-t^(2))hatk Distance are in meters, and the time t, in seconds. ( a) What is the velocity vector at t = + 3 ? (b) What is the speed ("in" m//"sec") at. t = + 3 ? (c) What is the acceleration vector and what is its magnitude ("in m"//"sec"^(2)) at.t = +3 ? (ii) Now the particle is moving only along the z-axis, and its position is given by (t^(2)- 2t -3)hatk k at what time does the particle stand still? _

A particle is moving with initial velocity 1m/s and acceleration a=(4t+3)m/s^(2) . Find velocity of particle at t=2sec .

A particle is moving along straight line whose position x at time t is described by x = t^(3) - t^(2) where x is in meters and t is in seconds . Then the average acceleration from t = 2 sec. to t = 4 sec, is :

A particle moves along the x-axis so that its position is given x = 2t^(3) –3t^(2) at a time t second. What is the time interval duringwhich particle will be on the negative half of the axis?

A particle moves in a straight line and its position x at time t is given by x^(2)=2+t . Its acceleration is given by :-