The momentum p of a particle changes with time according to the relation- dpdt-10-2t N- If the momentum is zero at t - 0 s- What will the momentum in Ns at t-10 s

Dpdt=10+2t ∫ dp=∫ (10+2t)dt p=10t+2t^22+c p=10t+t^2+c At t = 0; p=0 ⇒ c=0 p = 10t+ t^2 p at t = 10 s p=10× 10+10× 10=200 Ns ...

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The momentum $p$ of a particle changes with time according to the relation: $\frac{d p}{d t} = (10 + 2 t) N$ . If the momentum is zero at $t = 0 s$ . What will the momentum in $N s$ at $t = 10 s$

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Solution

$\frac{d p}{d t} = 10 + 2 t$
$\int d p = \int (10 + 2 t) d t$
$p = 10 t + \frac{2 t^{2}}{2} + c$
$p = 10 t + t^{2} + c$
At $t = 0; p = 0 \Rightarrow c = 0$
$p = 10 t + t^{2}$
$p at t = 10 s$
$p = 10 \times 10 + 10 \times 10 = 200 N s$

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The momentum p of a particle changes with time according to the relation: dpdt=(10+2t) N. If the momentum is zero at t=0 s. What will the momentum in Ns at t=10 s

dpdt=10+2t ∫dp=∫(10+2t)dt p=10t+2t22+c p=10t+t2+c At t=0;p=0⇒c=0 p=10t+t2 p at t = 10 s p=10×10+10×10=200 Ns

The momentum $p$ of a particle changes with time according to the relation: $\frac{d p}{d t} = (10 + 2 t) N$ . If the momentum is zero at $t = 0 s$ . What will the momentum in $N s$ at $t = 10 s$

$\frac{d p}{d t} = 10 + 2 t$
$\int d p = \int (10 + 2 t) d t$
$p = 10 t + \frac{2 t^{2}}{2} + c$
$p = 10 t + t^{2} + c$
At $t = 0; p = 0 \Rightarrow c = 0$
$p = 10 t + t^{2}$
$p at t = 10 s$
$p = 10 \times 10 + 10 \times 10 = 200 N s$