Question

a motorboat starting from rest on a lake acceleration in a straight line at a constant rate of 3.0ms for 8s. how does the boat travel during this time.​

Answer 1

u= 0 m/s

a= 3.0 m/s

t= 8s

by using, s= ut + 1/2at2

= 0t + 1/2 × 3×8×8

= 96 m

so, the distance covered by the boat is 96 m

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Distance\:travell=96\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Acceleration(a) = 3 \: {m/s}^{2} \\ \\ \tt: \implies Time(t) = 8 \: sec \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Distance(s) =?$

• According to given question :

$\tt \circ \: Initial \: velocity = 0 \: m/s \\ \\ \tt \circ \:Acceleration = 3 { \: ms}^{2} \\ \\ \tt \circ \: Time =8 \: sec \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} {at}^{2} \\ \\ \tt: \implies s = 0 \times 8 + \frac{1}{2} \times 3 \times {8}^{2} \\ \\ \tt: \implies s = 0 + \frac{1}{2} \times 3 \times 64 \\ \\ \green{\tt: \implies s = 96 \: m} \\ \\ \green{\tt \therefore Distance \: travelled \: in \: 8 \: sec \: is \: 96 \: m} \\ \\ \blue{ \bold{Some \: related \: formula}} \\ \orange{\tt \circ \: v = u + at} \\ \\ \orange{\tt \circ \: {v}^{2} = {u}^{2} + 2as}$