Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds.

The object is in motion, so its energy = kinetic energy = 1 /2 mv2  The kinetic energy of the object, when it comes to rest = 0.  Work done on object...

 The slope of the distance-time graph for an object in uniform motion is straight line.   The slope of the distance-time graph for an object in non-uniform motion is not...

(i) A body is said in uniform acceleration when its motion is along a straight line and its velocity changes by equal magnitude in equal interval of time.  (ii) A body...

No. Though the moving object comes back to its initial position the distance travelled is not zero. 

ω0 = 0 ; ρ = 100 rev/s ; ω = 2 ; ρ = 200 π rad/s ω = ω0 = αt ω = αt α= (200 π)/4 = 50 π rad /s2 or 25 rev/s2 θ = ω0t + 1/2...

(3) t = ax2 + bx by differentiating acceleration = - 2av3

θ = 100 π; t = 5 sec θ = 1/2 αt2 = 100 π = 1/2 α 25 α = 8π × 5 = 40 π rad/s = 20 rev/s α = 8π rad/s2 = 4 rev/s2 ω = 40π rad/s2...

θ = 5 rev, α = 2 rev/s2, ω0 = 0 ; ω = ? ω2 = (2 α θ) ω = √(2 x 2x 5) = 2 √5 rev/s. or θ = 10π rad,  α= 4π rad/s2, ω0 =...

(c) They will be together at the starting point after the L.C.M of 36, 48 and 42 L.C.M. of 36, 48, 42 = 1008 seconds

(c)  2/v = 1/v1 + 1/v2. Explanation: Time taken during first phase =  x/v1 and during second phase =  x/v2.  Total distance =2x   Total time taken = (x/v1)+(x/v2) = x(1/v1 +1/v2)     Average velocity =Total distance/Total...