Question

a car attains a velocity of 15 metre per second with in 5 second from 5 metre per second calculate the acceleration and displacement in this time internal?​

Answer 1

Answer:

$\bigstar{\bold{Acceleration=2\:m/s^{2} }}$

$\bigstar{\bold{Distance\:covered=50\:m}}$

Explanation:

$\Large{\underline{\rm{Given:}}}$

• Initial velocity = 15 m/s
• Final velocity = 5 m/s
• Time taken = 5 s

$\Large{\underline{\rm{To\:Find:}}}$

• Acceleration
• Displacement

$\Large{\underline{\rm{Solution:}}}$

Acceleration:

➔ First we have to find the acceleration of the car.

➔ By the first equation of motion we know that,

    v = u + at

   where v = final velocity

    u = initial velocity

    a = acceleration

    t = time taken

➔ Substitute the data,

    15 = 5 + a × 5

     5a = 15 - 5

     5a = 10

       a = 10/5

      a = 2 m/s²

➔ Hence acceleration of the car is 2 m/s².

    $\boxed{\bold{Acceleration=2\:m/s^{2} }}$

Displacement:

➔ Now we have to find the displacement of the car.

➔ By the third equation of motion we know that,

    v² - u² = 2as

    where v = final velocity

    u = initial velocity

    a = acceleration

    s = displacement

➔ Substitute the data,

    15² - 5² = 2 × 2 × s

    225 - 25 = 4s

    4s = 200

      s = 200/4

      s = 50 m

➔ Hence the distance covered by the car is 50 m.

    $\boxed{\bold{Distance\:covered=50\:m}}$

$\Large{\underline{\rm{Notes:}}}$

The three equations of motion are:

• v = u + at
• s = ut + 1/2 × a × t²
• v² - u² = 2as

Answer 2

Explanation:

Given:-

intial velocity =u=5m/s

final velocity =v=15m/s

time taken =t=5s

To find:-

Acceleration=a

Distance =s

Solution:-

• According to first equation of kinematics

${:}\longrightarrow$${\boxed{v=u+at}}$

${:}\longrightarrow$$15=5+a×5$

${:}\longrightarrow$$15-5=5a$

${:}\longrightarrow$$5a=10$

${:}\longrightarrow$$a={\frac {10}{5}}$

${:}\longrightarrow$${\underline{\boxed{\bf {2m/s {}^{2}}}}}$

• According to second equation of kinematics

${:}\longrightarrow$${\boxed {s=ut+{\frac {1}{2}}at {}^{2}}}$

${:}\longrightarrow$$s=5×5+{\frac {1 }{{\cancel{2}}}}×{\cancel{2}}×{5}^{2}$

${:}\longrightarrow$$s=25+25$

${:}\longrightarrow$${\underline{\boxed{\bf {s=50m}}}}$