Question

A particle is moving along a straight line such that
its velocity varies with position as shown in figure,
then the acceleration of the particle at x = 10 m is

(1) -4.6 m/s2
(3) -8.9 m/s2
(2) -6.8 m/s2
(4) -10.6 m/s2​

Answer 1

Answer:

Option (3) $-8.9$ m/s²

Explanation:

The equation fo the line shown in the figuure will be

$v-20=\frac{0-20}{15-0} (x-0)$

or, $v-20=\frac{-4}{3} x$

or, $3v-60=-4x$

or, $3v+4x-60=0$ ............(1)

This is the relation between v and x

At x = 10 m

∴ $3v+4\times 10-60=0$

$\implies 3v+40-60=0$

$\implies v=20/3$

We know that rate of change of velocity is acceleration and rate of change of displacement is velocity

Therefore, with little re-adjustment acceleration can be written in terms of v and x as

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

From eq (1)

$3v=60-4x$

Differentiating w.r.t. x we get

$3\frac{dv}{dx}=0-4$

$\implies \frac{dv}{dx}=\frac{-4}{3}$

At x = 10 m

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

$\implies a=\frac{20}{3}\times (-\frac{4}{3})$

$\implies a=-\frac{80}{9}$

$\implies a=-8.88$ m/s²

or, $a=-8.9$ m/s²

Hope this helps.

Answer 2

Answer:

Explanation:

answer