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Standard XI

Physics

Motion Under Variable Acceleration

A particle is...

Question

A particle is moving along the path given by y = $\frac{c}{6}$ $t^{6}$ (where c is positive constant). The relation between the acceleration (a) and the velocity (v) of the particle at t=5 sec. Is

$5 a = v$

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a = $\sqrt{v}$

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$a = 5 v$

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$a = v$

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Solution

The correct option is D
$a = v$

Differentiate y we get $v = c (t^{5})$
Differentiate v we get $a = 5 c (t^{4})$

In expressions for v and a, substitute t = 5 s and take ratio of the two quantities.

We get, v=a

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A particle is moving along the path given by y = c6 t6(where c is positive constant). The relation between the acceleration (a) and the velocity (v) of the particle at t=5 sec. Is

The correct option is D a=v Differentiate y we get v=c(t5) Differentiate v we get a=5c(t4) In expressions for v and a, substitute t = 5 s and take ratio of the two quantities. We get, v=a

A particle is moving along the path given by y = $\frac{c}{6}$ $t^{6}$ (where c is positive constant). The relation between the acceleration (a) and the velocity (v) of the particle at t=5 sec. Is

The correct option is D
$a = v$

Differentiate y we get $v = c (t^{5})$
Differentiate v we get $a = 5 c (t^{4})$

In expressions for v and a, substitute t = 5 s and take ratio of the two quantities.

We get, v=a