Question

A car moving with initial velocity of um/s. ofter applying the breaks,it retardation is 0.5m/s and it stop after 20sec . find the initial velocity and distance traveled by the car . ofter applying breaks​

Answer 1

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

$\red{➤}\:$$\textsf{Retardation of car }$$\sf = -0.5 m/s^2$

$\red{➤}\:$$\textsf{Time period before stopping}$$\sf = 20\: s$

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

$\orange{☛}\:$$\textsf{Initial Velocity of the car,u}$$\sf$

$\orange{☛}\:$$\textsf{Distance travelled by the car before }$

$\sf stopping,S$

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

$\underline{\tt{Formula\:Applied}}$

By first equation of motion-

$\green{ \underline { \boxed{ \sf{v-u=at}}}}$

where

• v = final velocity
• u = initial velocity
• a = acceleration
• t = time period

$\underline{\tt\pink{Putting \:Values-}}$

$\begin{gathered}\\\quad\longrightarrow\quad \sf 0-u = -0.5\times 20 \quad(v=0; body\: is\: finally\: at\:rest)\\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad \sf \cancel{-}u = \cancel{-}10\\\end{gathered}\quad$

$\begin{gathered}\\\quad\longrightarrow\quad \boxed{\sf {u = 10m/s}}\\\end{gathered}$

$\huge\orange{\dag}\:$$\textbf{Initial Velocity of the car is}$$\bf 10\:m/s$

By third equation of motion-

$\purple{ \underline{\boxed{ \sf{v^2-u^2=2aS}}}}$

• S = distance travelled

$\underline{\tt\pink{Putting \:Values-}}$

$\begin{gathered}\\\quad\longrightarrow\quad \sf 0^2-{10}^2 = 2\times -0.5 \times S \\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad \sf 0-100 = -1\times S \\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad \sf -100 = -S\\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad \boxed{\sf{S=100 \: m}} \\\end{gathered}$

$\huge\red{\dag}\:$$\textbf{Distance travelled by the car }$

$\bf is \:100\: metres$

Note-

=> Acceleration is taken negetive in case of retardation.

=> We can also find Distance travelled by second equation of motion-

$\green{ \underline{ \boxed{ \sf{S=ut+\dfrac{1}{2}at^2}}}}$

Answer 2

Answer:

Explanation:

Given,

• Retardation of the car, a = - 0.5 m/s²
• Time taken by the car, t = 20 seconds.

• Final velocity of the car, v = 0 m/s (As brakes applied)

To Find,

• Initial velocity of the car, u =?
• Distance traveled by car, s =?

Formula to be used,

• 1st equation of motion, v = u + at

• 3rd equation of motion, v² - u² = 2as

Solution,

Putting all the values, we get

v = u + at

⇒ 0 = u + (- 0.5) × 20

⇒ - u = - 0.5 × 20

⇒ u = 10 m/s

Hence, the initial velocity of the car is 10 m/s.

Now, the distance covered,

v² - u² = 2as

⇒ (0)² - (10)² = 2 × (- 0.5) × s

⇒ - 100 = - 1s

⇒ s = 100 m

Hence, the distance covered is 100 m.