A car starting from rest is accelerated at constant rate until it attains a constant speed, $$v$$. It is then retarded at a constant rate until it comes to rest. Considering that the car moves with constant speed for half of the time of total journey, the average speed of the car for the journey is
$$V_{avg} = \dfrac{ distance}{time}$$
Let the total time be $$t$$.
Average speed for 1st interval of $$\dfrac{t}{4}$$ is $$\dfrac{v}{2}$$
For 1st interval, $$s_1 = \dfrac{v}{2} \times \dfrac{t}{2}$$
Average speed for $$3rd$$ interval of $$\dfrac{t}{4}$$ is $$\dfrac{v}{2}$$
For $$3rd$$ interval, $$s_3 = \dfrac{v}{2} \times \dfrac{t}{2}$$
For $$2nd$$ interval, $$s_2 = v \times \dfrac{t}{2}$$
Total distance $$d = s_1 + s_2 + s_3$$
Total time is $$t$$
$$v_{avg} = \dfrac{d}{t}= \dfrac{3v}{4}$$
A particle starts from east and travels a distance $$S$$ with uniform acceleration, then it travels a distance $$2S$$ with uniform speed, finally it travels a distance $$3S$$ with uniform retardation and comes to rest. If the complete motion of the particle is a straight line then the ratio of its average to maximum velocity is:
A body moves with uniform velocity of $$u=7\;m/s$$ from $$t=0\;to\;t=1.5 s$$. It starts moving with an acceleration of $$10\;m/s^{2}$$. The distance between $$t=0\;to\;t=3s$$ will be:
A car starts moving rectilinearly, first with acceleration $$w=5.0\:m/s^2$$ (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate $$w$$, comes to a stop. The total time of motion equals $$\tau=25\:s$$. The average velocity during that time is equal to $$\left