Physics, asked by karan112233, 9 months ago

the speed of a car is reduced from 90 km per hour to 36 kilometre per hour in 5 seconds what is the distance travelled by the car during this time interval ​

Answers

Answered by ShivamKashyap08
146

Answer:

  • The car will travel a Distance (s) of 87.5 meters.

Given:

  1. Initial velocity (u) = 90 Km / h
  2. Final velocity (v) = 36 Km / h
  3. Time  (t) = 5 seconds.

Explanation:

\rule{300}{1.5}

From Conversion factor we know,

⇒ 1 Km/h = 5 / 18 m/s

⇒ 90 Km/h = 90 × 5 / 18

⇒ 90 Km/h = 5 × 5

u = 90 Km/h = 25 m/s

Similarly,

⇒ 1 Km/h = 5 / 18 m/s

⇒ 36 Km/h = 36 × 5 / 18

⇒ 36 Km/h = 2 × 5

v = 36 Km/h = 10 m/s.

\rule{300}{1.5}

\rule{300}{1.5}

Firstly finding the acceleration.

⇒ a = ( v - u ) / t

Where,

  • a Denotes Acceleration.
  • v Denotes Final velocity.
  • u Denotes Initial velocity.
  • t Denotes Time taken.

Now,

⇒ a = ( v - u ) / t

Substituting the values,

⇒ a = ( 10 - 25 ) / 5

⇒ a = - 15 / 5

⇒ a = - 3

a = - 3 m/s²

We got the acceleration.

\rule{300}{1.5}

\rule{300}{1.5}

From second Kinematic Equation we know,

s = u t + 1 / 2 a t²

Where,

  • a Denotes Acceleration.
  • s Denotes Distance travelled
  • u Denotes Initial velocity.
  • t Denotes Time taken.

Now,

⇒ s = u t + 1 / 2 a t²

Substituting the values,

⇒ s = 25 × 5 + 1 / 2 × - 3 × (5)²

⇒ s = 125 + 1 / 2 × - 3 × 25

⇒ s = 125 + 1 / 2 × - 75

⇒ s = 125 - 37.5

⇒ s = 87.5

s = 87.5 m.

The car will travel a Distance (s) of 87.5 meters.

\rule{300}{1.5}

Answered by StarrySoul
100

Given :

• Inital Velocity(u) = 90 km/hr

• Final Velocity(v) = 36 km/hr

• Time(t) = 5 seconds

To Find :

• Distance Travelled by the car during this interval

Solution :

First of all convert the given velocities of km/hr into m/sec because the time given is in seconds i.e.

 \star \sf \: u = 90 \times  \dfrac{5}{18}  = 25 \: m {s}^{ - 1}

 \star \sf \: v= 36\times  \dfrac{5}{18}  = 10\: m {s}^{ - 1}

Now, Let's Find Acceleration :

 \bigstar \large \:  \:  \:  \:  \:  \boxed{ \purple{ \sf \: a =  \frac{v - u}{t} }}  \:

 \longrightarrow \sf \: a =  \dfrac{10 - 25}{5}

 \longrightarrow \sf \: a =   \cancel\dfrac{ - 15}{5}

 \longrightarrow \sf \: a =   \cancel\dfrac{ - 15}{5}

 \longrightarrow \sf \: a =   - 3m {s}^{ - 2}

We've to Find the Distance now :

From 3rd equation of motion :

 \bigstar \large \:  \:  \:  \:  \:  \boxed{ \purple{ \sf \: 2as=   {v}^{2}  -  {u}^{2}  }}  \:

  \longrightarrow \sf \: 2( - 3)(s) =  ({10})^{2}  -  ({25})^{2}

  \longrightarrow \sf \:  - 6s=  100-  625

  \longrightarrow \sf \:  - 6s=   - 525

  \longrightarrow \sf \:  s=     \cancel\dfrac{ - 525}{ - 6}

  \longrightarrow \sf \:  s=  87.5 \: m

Hence,Distance Travelled by the car during this interval is 87.5 m

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