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 min
v (m/s)
20
(1) 4.6 m/s?
(3) -8.9 m/s?
(2) 6.8 m/s?
(4) -10.6 m/s​

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:

Option (3) -8.9−8.9 m/s²

Explanation:

The equation fo the line shown in the figuure will be

v-20=\frac{0-20}{15-0} (x-0)v−20=

15−0

0−20

(x−0)

or, v-20=\frac{-4}{3} xv−20=

3

−4

x

or, 3v-60=-4x3v−60=−4x

or, 3v+4x-60=03v+4x−60=0 ............(1)

This is the relation between v and x

At x = 10 m

∴ 3v+4\times 10-60=03v+4×10−60=0

\implies 3v+40-60=0⟹3v+40−60=0

\implies v=20/3⟹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}a=v

dx

dv

From eq (1)

3v=60-4x3v=60−4x

Differentiating w.r.t. x we get

3\frac{dv}{dx}=0-43

dx

dv

=0−4

\implies \frac{dv}{dx}=\frac{-4}{3}⟹

dx

dv

=

3

−4

At x = 10 m

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

dx

dv

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

3

20

×(−

3

4

)

\implies a=-\frac{80}{9}⟹a=−

9

80

\implies a=-8.88⟹a=−8.88 m/s²

or, a=-8.9a=−8.9 m/s²