A particle starts from rest and its angular displacement in rad is given by - - t220 - t5- If the angular velocity at the end of t - 4 s is -k-Then- what will be the value -5k- -

Displacement θ = t ^ 2 20 + t 5 Angular velocity, #160; ω= dθ #160; dt = 2t 20 + 1 5 #160; #160; #160; #160; #160; #160; ...

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Standard XII

Mathematics

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A particle starts from rest and its angular displacement (in rad) is given by $θ = \frac{t^{2}}{20} + \frac{t}{5}$ . If the angular velocity at the end of $t = 4 s$ is $^{'} k^{'}$ .Then, what will be the value $^{'} 5 k^{'}$ ?

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Solution

Displacement $θ = \frac{t^{2}}{20} + \frac{t}{5}$

Angular velocity,
$ω = \frac{d θ}{d t} = \frac{2 t}{20} + \frac{1}{5}$
$= \frac{t}{10} + \frac{1}{5}$

$\frac{d θ}{d t} |_{t = 4} = \frac{4}{10} + \frac{1}{5} = 0.4 + 0.2 = 0.6$

$∴ k = 0.6$

$∴ 5 k = 0.6 \times 5 = 3$

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A particle starts from rest and its angular displacement (in rad) is given by θ=t220+t5. If the angular velocity at the end of t=4 s is ′k′.Then, what will be the value ′5k′ ?

Displacement θ=t220+t5Angular velocity, ω=dθdt=2t20+15 =t10+15dθdt|t=4=410+15=0.4+0.2=0.6∴k=0.6∴5k=0.6×5=3

A particle starts from rest and its angular displacement (in rad) is given by $θ = \frac{t^{2}}{20} + \frac{t}{5}$ . If the angular velocity at the end of $t = 4 s$ is $^{'} k^{'}$ .Then, what will be the value $^{'} 5 k^{'}$ ?

Displacement $θ = \frac{t^{2}}{20} + \frac{t}{5}$

Angular velocity,
$ω = \frac{d θ}{d t} = \frac{2 t}{20} + \frac{1}{5}$
$= \frac{t}{10} + \frac{1}{5}$

$\frac{d θ}{d t} |_{t = 4} = \frac{4}{10} + \frac{1}{5} = 0.4 + 0.2 = 0.6$

$∴ k = 0.6$

$∴ 5 k = 0.6 \times 5 = 3$