Question

16. For a body travelling with uniform accela
ration, its final velocity is v=
$\sqrt{180 - 7x}$

where x is the distance travelled by the bo
Then the acceleration is
2) -3.5 m/s2
3) -7 m/s2
4) 180 m/s2
1) -8 m/s2
TE 1


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Answer 1

Answer:

$\boxed{\mathfrak{(2) \ -3.5 \ m/s^2}}$

Given:

Velocity (v) w.r.t distance travelled (x):

$\sf v = \sqrt{180 - 7x}$

To Find:

Acceleration (a)

Explanation:

Rate of change of velocity w.r.t time is equal to acceleration:

$\sf \implies a = \frac{dv}{dt} \\ \\ \sf \implies a = \frac{dv}{dt} \times \frac{dx}{dx} \\ \\ \sf \implies a = \frac{dx}{dt} \times \frac{dv}{dx} \\ \\ \sf \frac{dx}{dt} = v : \\ \sf \implies a =v \frac{dv}{dx} \\ \\ \sf v = \sqrt{180 - 7x} : \\ \sf \implies a = (\sqrt{180 - 7x} ) \times \frac{d( \sqrt{180 - 7x} )}{dx} \\ \\ \sf \implies a = \cancel{\sqrt{180 - 7x}} \times \frac{ - 7}{2 \cancel{ \sqrt{180 - 7x} }} \\ \\ \sf \implies a = - \frac{7}{2} \\ \\ \sf \implies a = - 3.5 \: m/s^2$

$\therefore$

Acceleration (a) = -3.5 m/s²

Answer 2

Answer:

a = vdv/dx

v = √(180 - 7x)

a = √(180 - 7x) × d√(180 - 7x)/dx

a = √(180 - 7x) × -7/2√(180 - 7x)

a = -7/2

a = -3.5 m/s²