Question

slove:- If u= 8m/s, a= 6m/S2 find 'S' if v= 10m/s.​

Answer 1

Given:

• Initial velocity ( u ) = 8 m/s
• Acceleration ( a ) = 6 m/s²
• Final velocity ( v ) = 10 m/s

To find :

• Distance ( s ) = ?

Solution:

Firstly finding the time taken

Using. ,

a = v - u/t

Simplifying

t = (v - u)/a

Substituting the given values

→ t = 10 - 8/6

→ t = 2/6

→ t = 1/3

→ t ≈ 0.33 s

_______________________

Now , finding distance ( s ) by using third equation of motion

S = ut + 1/2 at²

→ S = 8(0.33) + 1/2(6)(0.33)²

→ S = 2.64 + 0.3267

→ S = 2.9667 m ≈ 3 m

Hence , distance travelled ≈ 3 m ( approx )

Answer 2

$\large \bf \red{ \underline { \underline{Given:-}}}$

• Initial velocity (u) = 8m/s

• Final velocity (v) = 10m/s

• Acceleration (a) = 6m/s²

$\large \bf \blue{ \underline { \underline{To\:Find:-}}}$

• The distance covered (s)


$\large \bf \green{ \underline { \underline{Solution:-}}}$

According to the first equation of motion

$\large \boxed{ \sf \pink{ { \rightarrow v = u + at}}}$

Substituting the values:-

$\sf\rightarrow 10= 8 + 6(t)$

$\sf\rightarrow 10 - 8 = 6(t)$

$\sf\rightarrow \dfrac{2}{6} = t$

$\sf\rightarrow t = 0.33sec$

According to the second equation of motion

$\boxed{ \sf \orange{\rightarrow s = ut + \frac{1}{2} a {t}^{2} }}$

Substituting the values:-

$\sf \rightarrow s = 8 \times 0.33 + \dfrac{1}{2} \times 6 \times {(0.33)}^{2}$

$\sf \rightarrow s = 2.64 + 0.3267$

$\sf \rightarrow s = 2.9667 = 3 approx$

Distance = 3m ..... approx