Question

A partile movint in a straight line with a velocity v(t)=2t^2 how far does the particle move between times t=1 and t=3

Answer 1

velocity is given as a function of time as v(t)=2t^2

thus on integrating velocity with limits from 1 and 3 we get s(t)=(2/3)t^3

putting limits and subtracting  the higher limit from lower

s(3)=(2/3)3^3=162

s(1)=(2/3)1^3=0.67

s(3)-s(1)=161.33 units

thus in 1 s to 3 s it displaced 161.33 units

Answer 2

$\underline{\underline{\textsf{Solution :}}}$

$\textsf{The velocity of the particle is}$

$\textsf{v(t) =} \textsf{2}{\textsf{t}}^{\textsf{2}}$

⇒ $\dfrac{\textsf{dx}}{\textsf{dt}} \textsf{= 2}{\textsf{t}}^{\textsf{2}}$

$\textsf{where x is the required distance}$
$\textsf{travelled in time t}$

⇒ dx = 2t² dt

$\textsf{Integrating, we get}$

∫ dx = ∫ 2t² dt

⇒ x = $\dfrac{2}{3}{\text{t}}^{3}$

$\textsf{Thus, the required distance}$
$\textsf{travelled between t = 1 and t = 3 be}$

$\textsf{x ( t = 1, t = 3 )}$

= $\dfrac{2}{3} * {\text{t}}^{3}$ ( t = 1, t = 3 ) units

= $\dfrac{2}{3} * (3^{3} - 1^{3})$ units

= $\dfrac{2}{3} * (27 - 1)$ units

= $\dfrac{2*26}{3}$ units

= $\dfrac{52}{3}$ units

= $\underline{\textsf{17.3 units ( approximately )}}$