The acceleration of particle varies with time as shown.Calculate the displacement of the particlein m in the interval from $$t=2$$ s to $$t=4$$ s. Assume that $$v=0$$ at $$t=0$$.

$$a=2t-2$$ $$(from$$ the $$graph)Now$$, ∫$$0vdv=$$∫$$0tadt=$$∫$$0t(2t-2)dt$$∴ $$v=t2-2tdsdt=t2-2t$$ ⇒∫$$0s$$ $$ds=$$∫$$24t2ddt-$$∫$$242tdt$$ ⇒$$s=203$$ m

Rectilinear Motion

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The acceleration of particle varies with time as shown.
(a) Find an expression for velocity in terms of t. (b) Calculate the displacement of the ..."> (a) Find an expression for velocity in terms of t. (b) Calculate the displacement of the ...">

Calculate the displacement of the particle $(i n m)$ in the interval from $t = 2 s$ to $t = 4 s$ . Assume that $v = 0$ at $t = 0$ .

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Solution

$a = 2 t - 2$ (from the graph)
Now, $\int_{0}^{v} d v = \int_{0}^{t} a d t = \int_{0}^{t} (2 t - 2) d t$
$∴ v = t^{2} - 2 t$
$\frac{d s}{d t} = t^{2} - 2 t \Rightarrow \int_{0}^{s} d s = \int_{2}^{4} t^{2} d d t - \int_{2}^{4} 2 t d t \Rightarrow s = \frac{20}{3} m$

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Rectilinear Motion

Remember concepts with our Masterclasses.

a=2t-2 (from the graph)Now, ∫0vdv=∫0tadt=∫0t(2t-2)dt∴ v=t2-2tdsdt=t2-2t ⇒∫0s ds=∫24t2ddt-∫242tdt ⇒s=203 m

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$a = 2 t - 2$ (from the graph)
Now, $\int_{0}^{v} d v = \int_{0}^{t} a d t = \int_{0}^{t} (2 t - 2) d t$
$∴ v = t^{2} - 2 t$
$\frac{d s}{d t} = t^{2} - 2 t \Rightarrow \int_{0}^{s} d s = \int_{2}^{4} t^{2} d d t - \int_{2}^{4} 2 t d t \Rightarrow s = \frac{20}{3} m$