Question

A particle moves in a straight line from a point with an initial velocity 10m/s and an retardation of 2m/s^2. Find the displacement and velocity of the particle after (i) t=5sec & (ii) t=10sec.

Friends pls solve this sum...it is urgent... ​

Answer 1

$\boxed{\mathsf{\green{Answer}}}$

At t = 5sec

velocity = 0 m/s

Displacement = 25m

At t = 10sec

velocity = -10m/s

Displacement = 0m

$\boxed{\mathsf{\green{Explanation}}}$

Given:

• u = 10m/s
• a = -2m/s

To find:

Displacement and velocity after

(i) t = 5sec

(ii) t = 10sec

Solution

At t = 5sec

$\boxed{\mathsf{\blue{v = u + at}}}$

$\mathsf{v = 10 + (-2)(5)}$

$\mathsf{v = 10 -10}$

$\mathsf{v = 0}$

$\boxed{\mathsf{\blue{S = ut + \dfrac{1}{2}a{t}^{2}}}}$

$\mathsf{S = (10)(5) + \dfrac{1}{2}(-2)({5})^{2}}$

$\mathsf{S = 50-25}}$

$\mathsf{S = 25m}$

At t = 10sec

$\boxed{\mathsf{\pink{v = u + at}}}$

$\mathsf{v = 10 + (-2)(10)}$

$\mathsf{v = 10 -20}$

$\mathsf{v = -10 m{s}^{-1}}$

$\boxed{\mathsf{\pink{S = ut + \dfrac{1}{2}a{t}^{2}}}}$

$\mathsf{S = (10)(10) + \dfrac{1}{2}(-2)({10})^{2}}$

$\mathsf{S = 100 -100}$

$\mathsf{S = 0m}$

$\texttt{\blue{Aravind}\: \red{Reddy}....!}$

Answer 2

$\bf{\huge{\boxed{\underline{\mathbb{\green{Answer:}}}}}}$

$\sf{at \: time = 5sec}$

Velocity = 0 m/s

Displacement = 25 m

$\sf{at \: time = 10 sec}$

$\huge \tt{given \: - > }$

★ u = 10m/s

★ a = -2m/s

NOW , WE HAVE TO FIND DISPLACEMENT AND VELOCITY AFTER AT :-

¡) time (t) = 5 sec

¡¡) time (t) = 10 sec

$\huge \tt{solution - > }$

$\implies\sf{at \: time = 5sec}$

BY USING FORMULA →

$\tt{\boxed{{\mathbb{\green{v = u \: + \: a \: t}}}}}$

$\tt{v = 10 + ( - 2)(5)} \\ \tt{v = 10 - 10} \\ \tt{v = 0 \: m {s}^{ - 1} }$

$\tt{\boxed{{\mathbb{\green{s = u \: t \: + \frac{1}{2} \: a \: {t}^{2} }}}}}$

$\tt{s \: = (10)(5) + \frac{1}{2} ( - 2) ({5})^{2} }$

$\tt{s = 50 - 25}$

$\tt{s = 25m}$

$\implies\sf{at \: time = 10 sec}$

$\tt{\boxed{{\mathbb{\green{v = u \: + \: a \: t}}}}}$

$\tt{v = 10 + ( - 2)(10)} \\ \tt{v = 10 - 20} \\ \tt{v = - 10 \: m {s}^{ - 1} }$

$\tt{\boxed{{\mathbb{\green{s = u \: t \: + \frac{1}{2} \: a \: {t}^{2} }}}}}$

$\tt{s = (10)(10) + \frac{1}{2} ( - 2)( {10})^{2} } \\ \\ \tt{s = 100 - 100} \\ \\ \tt{s = 0 \: m}$