Question

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 ​

Answer 1

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.

$\rule{300}{1.5}$

Answer 2

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