Subjective Type

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 $$\left=72\:km/h$$. The car moves uniformly for (10+x) seconds. The value of x is:
Solution
As the car starts from rest and finally comes to a stop, and the rate of acceleration and deceleration are equal, the distances as well as the times taken are same in these phases of motion.
Let $$\Delta t$$ be the time for which the car moves uniformly. Then the acceleration/deceleration time is $$\displaystyle\dfrac{\tau-\Delta t}{2}$$ each. So total distance is calculated as
$$s=2(ut+\dfrac{1}{2}at^2)+v_{max}t$$
$$v_{max}$$ is calculated using $$v=u+at=0+w\dfrac{\tau-\Delta t}{2}$$
Thus we get
$$\displaystyle\left\tau=2\left\{\dfrac{1}{2}w\frac{{(\tau-\Delta t)}^2}{4}\right\}+w\frac{(\tau-\Delta t)}{2}\Delta t$$
$$=w\dfrac{\tau-\Delta t}{2}\left(\dfrac{\tau-\Delta t}{2}+\Delta t\right) $$
$$=w\dfrac{\tau-\Delta t}{2}\left(\dfrac{\tau+\Delta t}{2}\right)$$
Solving it we get
$$\displaystyle\Delta t^2=\tau^2-\frac{4\left\tau}{w}$$
Hence $$\displaystyle\Delta t=\tau\sqrt{1-\dfrac{4\left}{w\tau}}=15\:s$$.
Kinematics
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
Kinematics
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:
Kinematics
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: