Question

The displacement x of a particle moving in one dimension is related to time t by the relation x = √(2t^2 - 3t), where x is in metres and t is in seconds. Fine the displacement of the particle when velocity is 0

Answer 1

Answer:

3/2√2 m

Explanation:

we know,

v=dx/dt

so,

d(√2t²-3t)/dt

applying chain rule,

d(√2t²-3t)/d(2t²-3t)*d(t)

=>1/2√(2t²-3t)*(4t-3)

so,

as v=0

1/2√(2t²-3t)*(4t-3)=0

so,4t-3=0 as denominator can't be zero

t=3/4 s

so at t=3/4 s ,velocity is 0

so,displacement at 3/4s=√2t²-3t

=√(2*(3/4)²-3*3/4)

=√(9/8-9/4)

=√9/8

=3/2√2 m

Answer 2

The correct answer is $\frac{3\sqrt{2} }{4}$.

Given: The equation of displacement = $x=\sqrt{2t^{2} -3t}$.

To Find: Displacement when velocity is zero.

Solution:

$x=\sqrt{2t^{2} -3t}$

For velocity differentiate this equation and equate it to zero.

$\frac{dx}{dt}=\frac{d(\sqrt{2t^{2}-3t})}{dt}$

Apply chain rule.

v = $\frac{1}{2\sqrt{2t^{2} -3t}} (4t-3)$

v = 0

$\frac{1}{2\sqrt{2t^{2} -3t}} (4t-3)$ = 0

Equate numerator to zero.

4t - 3 = 0

t = $\frac{3}{4}$

Now put in displacement equation.

$x=\sqrt{|2t^{2} -3t|}$

x = $\sqrt{|2(\frac{3}{4} )^{2} -3(\frac{3}{4} )|}$

x = $\sqrt{|(\frac{9}{8} ) -(\frac{9}{4} )|}$

x = $\frac{3}{2}\sqrt{|\frac{1}{2} -1|}$

x = $\frac{3}{2\sqrt{2} }$

Divide and multiply by $\sqrt{2}$ to rationalize.

x = $\frac{3}{2\sqrt{2} }*\frac{\sqrt{2} }{\sqrt{2} }$

x = $\frac{3\sqrt{2} }{4}$

Hence, the displacement of the particle when velocity is 0 is $\frac{3\sqrt{2} }{4}$.

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