A particle is moving with a velocity of $v = (3 + 6 t + 9 t^{2})$ cm/s. Find out the displacement of the particle in the interval $t = 5 s$ to $t = 8 s$ .

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Solution

Velocity of the particle $v = (3 + 6 t + 9 t^{2}) c m / s$

Displacement of the particle $d S = v d t$

Integrating $\int_{s_{1}}^{s_{2}} d S = \int_{5}^{8} (3 + 6 t + 9 t^{2}) d t$

$∴$ $S {∣ ∣ ∣}_{s_{1}}^{s_{2}} = 3 t {∣ ∣ ∣}_{5}^{8} + 6 \times \frac{t^{2}}{2} {∣ ∣ ∣}_{5}^{8} + 9 \times \frac{t^{3}}{3} {∣ ∣ ∣}_{5}^{8}$

OR $Δ S = S_{2} - S_{1} = 3 (8 - 5) + 3 (8^{2} - 5^{2}) + 3 (8^{3} - 5^{3}) = 1287 c m$

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A particle is moving with a velocity of v=(3+6t+9t2)cm/s. Find out the displacement of the particle in the interval t=5s to t=8s.

Velocity of the particle v=(3+6t+9t2)cm/s Displacement of the particle dS=vdtIntegrating ∫s2s1dS=∫85(3+6t+9t2)dt∴ S∣∣∣s2s1=3t∣∣∣85+6×t22∣∣∣85+9×t33∣∣∣85OR ΔS=S2−S1=3(8−5)+3(82−52)+3(83−53)=1287cm

A particle is moving with a velocity of $v = (3 + 6 t + 9 t^{2})$ cm/s. Find out the displacement of the particle in the interval $t = 5 s$ to $t = 8 s$ .