Question

A body is moving with a velocity 5 m/s and an acceleration of 10 m/$s^2$. Which of these are correct?

• A.
• B.
• C.
• D.

Answer 1

Answer: (A), (B), (C)

(a) u = 5 m/s a = 10 m/$s^2$

v = 5 + 10t so this is like y = mx + c equation of straight line.

(b) $s=ut+\frac{1}{2}a{t}^2$

s = 5t + 5$t^2$

y = $a{x}^2+bx$ parabolic equation

Substitute values and check and draw

t = 0, s = 0; t = 1s = 10, t = 2, s = 30

So we see the rate of increase of s is also increasing.Which should be the case as the body is

accelerating, hence velocity or $\frac{ds}{dt}$ is increasing.

So graph b is correct

(c) $v^2=u^2+2as$ $v^2=y$ $\Rightarrow y=25+2ax$

$v^2=25+2as$ $s=x$ $y=mx+c$

Straight line form. You can ever substitute some values and draw.

(d) From above we have $v^2=20x+25$

i.e., $y^2=20x+25\Rightarrow y^2=ax+c$

How will go draw $y^2=x$

x = 0 y = 0

If x = 1 y = 1

x = 4 y = 2

x = 9 y = 3

$y^2=x+c$ will just shift the graph a little bit up.

Which is not similar to the given option d.

Also $v^2=20s+25\to$ differentiating w.r.t.s we get

$\alpha \times \frac{dv}{ds}=20\Rightarrow \frac{dv}{ds}=\frac{10}{v}$ So since v increase the slope decreases.Which proves our graph is correct.