Correct Option is (C)

From given information :

For 1^st interval

$a_1=\frac{v_0}{t_1}$

v₀ = a₁ t₁ ....... (1)

For 2^nd interval

$a_2=\frac{v_0}{t_2}$

v₀ = a₂ t₂ ..... (2)

from (1) & (2)

a₁ t₁ = a₂ t₂

$\frac{t_1}{t_2}=\frac{a_2}{a_1}$

Correct Option is (C)

From the graph for first line, the slope is negative and intercept is positive.

So, equation of line is

$v=-mt+c$

$\Rightarrow a_1=\frac{dv}{dt}=-m$

Similarly, for second line, the slope is positive and intercept is negative, so equation of line is

$v=mt-c$

$\Rightarrow a_2=\frac{dv}{dt}=m$

$\therefore$ The corresponding acceleration-time graph as shown below


Hence, option (a) is correct.

Correct answer is (1)

y = mx + C

v² = $\frac{20}{10}$x + 20

v² = 2x + 20

2v$\frac{dv}{dx}$ = 2

$\therefore$ a = v$\frac{dv}{dx}$ = 1

Correct Option is (D)

4.333 sec = $\frac{13}{3}$ sec

Distance = area under the v-t graph

= Area of Parallelogram + Area of triangle

= $\frac{1}{2}\left(4\right)\left(\frac{13}{3}+1\right)$ + $\frac{1}{2}\left(6-\frac{13}{3}\right)\times 2$

= $\frac{37}{3}$ m

Correct Option is (C)

V, h curve will be parabolic.

Downward velocity is negative and upward is positive

When ball is coming down graph will be in IV quadrant and when going up graph will be in I quadrant.

At H = h, v = 0

at h = 0, v = $\sqrt{2gh}$

also a = –g, throughout this motion.