Question

A ball travelling in straight line with initial velocity 49m/s and acceleration -10 m/s find dost travelled in 5 Th second

Answer 1

$\huge{\bold{\underline{\underline{\tt Correct\: question:-}}}}$

A ball travelling in straight line with initial velocity 49m/s and acceleration -10 m/s find distance travelled in 5 th second?.

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Initial velocity (u) = 49 m/s.
• Acceleration (a) = - 10 m/s².
• Time taken (t) = 5 seconds.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

By applying second kinematic equation,

$\large{\boxed{\tt S = ut + \dfrac{1}{2}at^2}}$

Substituting the values,

$\large{\tt S = 49 \times 5 + \dfrac{1}{2} \times- 10 \times (5)^2}$

$\large{\tt S = 245 - \dfrac{1}{2} \times 10 \times 25}$

$\large{\tt S = 245 - \dfrac{1}{\cancel{2}} \times \cancel{10} \times 25}$

Now,

$\large{\tt S = 245 - 5 \times 25}$

$\large{\tt S = 245 - 125}$

$\huge{\boxed{\boxed{\tt S = 120 \: meters}}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Additional formulas:-

• $\large{\tt v^2 - u^2 = 2as}$
• $\large{\tt v = u + at}$
• $\large{\tt S = ut + \dfrac{1}{2}at^2}$

• $\large{\tt S_n = u + \dfrac{a}{2}(2n - 1)}$

$\rule{300}{1.5}$

Answer 2

Explanation:

$Helloo..... \\ Here \: given \: that = > \\ u = Initial \: velocity = 49m {s}^{ - 1} \\ a = Acceleration = - 10m {s}^{ - 1} \\ here \: - ve \: sign \: shows \: that \: it \: \\ has \: been \: retarded..... \\ To \: find = Distance \: travelled \: in \: 5th \: second. \\ \\ by \: newtons \: second \: law = > \\ \\ S = ut + \frac{1}{2} a {t}^{2} \\ S = 49(5) + \frac{1}{2} ( - 10)( {5})^{2} \\ S = 245 - 5(25) \\ S = 245 - 125 \\ S = 120 \: m \\ \\ \\ Hope \: it \: solution \: is \: clear$