Question

A body moving with an intially velocity of 5 m\s attains a velocity of 15 m\s in 3 s.then the displacement is​

Answer 1

Answer:

$\mathfrak{\huge{\underline{Answer:}}}$

Explanation:

Given:

Initial velocity:[U]: 5m/s

Final velocity [V]: 15 m/s

time [t]:3 sec

V=U+at

15=5+a*3

15-5=3a

a=10/3

a=3.33 m/s²

To find displacement

V²=U²+2aS .......[S= displacement]

15²=5²+2*3.33*S

225-25=6.66S

S=200/6.66

S=30.3m

Answer 2

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

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

• Initial velocity (u) = 5 m/s.
• Final velocity (v) = 15 m/s.
• Time taken (t) = 3 seconds.

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

Applying First kinematic equation,

$\large{\boxed{ \tt v = u + at}}$

Substituting the values,

$\large{ \tt 15 = 5 + a \times 3}$

Now,

$\large{ \tt 15 - 5 = 3a}$

$\large{ \tt 10 = 3a}$

$\large{ \tt a = \dfrac{10}{3}}$

$\large{\boxed{\tt a = 3.333 \: m/s^2}}$

Applying second kinematic equation,

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

Substituting the values,

$\large{\tt S = 5 \times 3 + \dfrac{1}{2} \times 3.333 \times (3)^2}$

$\large{\tt S = 15 + \dfrac{1}{2} \times 3.333 \times 9}$

$\large{\tt S = 15 + \dfrac{1}{2} \times 29.977}$

$\large{\tt S = 15 + 14.99}$

$\huge{\boxed{\boxed{\tt S = 29.99 \: m}}}$

So, the displacement is 29.99 meters or approximately 30 meters.

Important formulas:-

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