A particle moves along a path \(ABCD\) as shown in the figure. The magnitude of the displacement of the particle from \(A\) to \(D\) is:

       

1. $(5+10\sqrt{2})$m
2. \(10\) m
3. $15\sqrt{2}$ m
4. \(15\) m

The velocity of a particle moving along a straight line with constant acceleration \(a\) reduces to \(\frac{1}{5}\) of its initial velocity in time \(\tau.\) The total time taken by the body till its velocity becomes zero is:
1. \(\frac{4\tau}{3}\)
2. \(\frac{5\tau}{4}\)
3. \(\frac{4\tau}{5}\)
4. \(\frac{3\tau}{4}\)

A boy falls from a building of height \(320\) m. After \(5\) seconds, superman jumps downward with initial speed \(u\) such that the boy can be saved. The minimum value of \(u\) is: (assume \(g= 10~\text{m/s}^2\)${\mathrm{}}^{}$)

The displacement \(x\) of a body having relation with time as \(x = 3t^2 -6t+7,\) then distance covered by body in first two second is [where \(x\) is in metre and \(t\) is in second].
1. Zero

2. \(6\) m

3. \(3\) m

4. \(9\) m

The acceleration of particle for which the ${\mathrm{v}}^2$ versus s graph is shown is:
                 
1. - 8 $\mathrm{m}/{\mathrm{s}}^2$
2. 4 $\mathrm{m}/{\mathrm{s}}^2$
3. - 4 $\mathrm{m}/{\mathrm{s}}^2$
4. -16 $\mathrm{m}/{\mathrm{s}}^2$

The position (x) of a particle moving along x-axis varies with time (t) as $\mathrm{x}={\mathrm{t}}^2-4\mathrm{t}+5$. The particle turns around at

1. t=1 s

2. t=2 s

3. t=3 s

4. t=4 s