Question

Position of a particle moving along x-axis is given by x=2+8t-4t*2. the distance travelled by the particle from t=0 to t=2 is​

Answer 1

$\dag\:\underline{\sf AnsWer :}$

Here we are asked to find the distance travelled by the particle from t = 0 to t = 2.

$:\implies \sf Velocity = \dfrac{Displacement}{time}$

Differentiating w.r.t to t

$:\implies \sf v = \dfrac{dx}{dt}$

$:\implies \sf v = \dfrac{d}{dt}(2 + 8t - 4 {t}^{2})$

$\bigstar\:\sf Rule = \dfrac{dx}{dt} =n x^{n - 1}$

$:\implies \sf v(t) =0 + 8-2 \times 4 t$

$:\implies \sf v(t) = 8-8 t$

Now,

$\dashrightarrow\:\:\displaystyle \sf Distance = \int\limits_{t_{1}}^{t_{2}} v(t) \, dt$

$\dashrightarrow\:\:\displaystyle \sf Distance = \int\limits_{0}^{2} (8 - 8t) \, dt$

$\bigstar\:\sf Rule = \displaystyle \sf \int\limits {x}^{n} \, dx = \dfrac{ {x}^{n + 1} }{n + 1}$

$\dashrightarrow\:\:\displaystyle \sf Distance = \Bigg[ 8t -\dfrac{8 {t}^{1 + 1}}{1 + 1} \Bigg]_{0} ^{2}$

$\dashrightarrow\:\:\displaystyle \sf Distance = \Bigg[ 8t -\dfrac{8 {t}^{2}}{2} \Bigg] _{0} ^{2}$

$\dashrightarrow\:\:\displaystyle \sf Distance = \Bigg[ 8(2) -\dfrac{8 {(2)}^{2}}{2} - 0 \Bigg]$

$\dashrightarrow\:\:\displaystyle \sf Distance = \Bigg[ 16 -\dfrac{8 \times 4}{2} - 0 \Bigg]$

$\dashrightarrow\:\:\displaystyle \sf Distance = \Bigg[ 16 -\dfrac{16}{2} \Bigg]$

$\dashrightarrow\:\:\displaystyle \sf Distance = 16 -8$

$\dashrightarrow\:\: \underline{ \boxed{\displaystyle \sf Distance = 8 \: m}}$