Question

A car accelerates uniformly from 18 km/h to 36 km/h in 5 sec.Calculate the distance covered by the car in that time *


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Answer 1

Answer:

Distance Travelled in 5 seconds = 37.5 m or 0.0375 km

Given:

• Initial Velocity = 18 km/h

• Final Velocity = 36 km/h

• Time travelled = 5 seconds

Concept:

Distance(S) = $ut+\dfrac{1}{2}at^2$

1 km/h = $\dfrac{5}{18}$ m/sec

To Find:

Distance travelled in 5 seconds

Explanation:

First converting velocities from km/h into m/sec.

Initial Velocity = $\dfrac{5}{18}\times 18$ = 5 m/sec.

Final Velocity = $\dfrac{5}{18}\times 36$ = 10 m/sec.

We know that

acceleration a =$\dfrac{(v – u)}{t}$

Substituting the given values in the above equation we get,

a = $\dfrac{(10 – 5)}{5}$

a =$\dfrac{5}{5}$

a =1 m/$s^2$

We can calculate the distance travelled by the below formula:

Distance travelled $(S) = u t + \left(\dfrac{1}{2}\right) a \times t^2$

Substituting the given values in the above equation we get,

$S = 5 \times 5 + \dfrac{1}{2}\times1\times5^2$

$S = 25 + \dfrac{1}{2}\times 25$

S = 25 + 12.5

S = 37.5 m

Therefore,

Distance travelled =37.5 m or 0.0375

Other Important Formulas:

Equations Of Motion:

$\boxed{\tt 1) \: v = u + at}$

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

$\boxed{\tt 3) \: {v}^{2} - {u}^{2} = 2aS}$

where

u = initial velocity of the body

v = final velocity of the body

a = acceleration of the body

t = time taken for change

s = distance covered

g = acceleration due to gravity

In some cases , if acceleration is not given and acceleration due to gravity is given , then we substitute g instead of a.

$\sf v = u + gt$

$\sf S = ut + \dfrac{1}{2} {gt}^{2}$

$\sf {v}^{2} - {u}^{2} = 2gh$

If the body is going up against acceleration due to gravity, then:

$\sf v = u - gt$

$\sf S = ut - \dfrac{1}{2} {gt}^{2}$

$\sf {v}^{2} - {u}^{2} = -2gh$

Answer 2

→ To Find :

The distance covered by the car .

→ We Know :

First Equation of Motion :

$\blue{\sf{\underline{\boxed{v = u + at}}}}$

Where ,

• v = Final velocity
• u = Initial velocity
• t = Time taken
• a = Acceleration produced

Third Equation of Motion :

$\blue{\sf{\underline{\boxed{v^{2} = u^{2} + 2aS}}}}$

Where ,

• v = Final velocity
• u = Initial velocity
• S = Distance covered
• a = Acceleration produced

Concept :

According to the question , we have to find the Distance covered by the car .

So by using the Third Equation of Motion ,we can find the Distance .

But first we have to find the acceleration produced by the car, by using the First Equation of Motion.

Solution :

Given :

• Initial velocity (u) = 18 km/h
• Final velocity (v) = 36 km/h
• Time taken (t) = 5 sec

Converting the the velocity in m/s.

To Convert the velocity in m/s from km/h , we multiply it by 5/28.

• $\sf{18 kmh^{-1}}$

$\:\:\:\:\:\:\:\:\sf{\rightarrow \bigg(18 \times \dfrac{5}{18}\bigg)ms^{-1}}$

$\:\:\:\:\:\:\:\:\sf{\rightarrow 5 ms^{-1}}$

• $\sf{36 kmh^{-1}}$

$\:\:\:\:\:\:\:\:\sf{\rightarrow \bigg(36 \times \dfrac{5}{18}\bigg)ms^{-1}}$

$\:\:\:\:\:\:\:\:\sf{\rightarrow 10 ms^{-1}}$

Hence , we got the new values as :

• Initial velocity (u) = 5 m/s
• Final velocity (v) = 10m/s

Acceleration Produced :

• Initial velocity (u) = 5 m/s
• Final velocity (v) = 10m/s
• Time taken (t) = 5 sec

Using the First Equation of Motion , and substituting the values in it , we get :

$\purple{\sf{v = u + at}}$

$\sf{\Rightarrow 10 = 5 + a \times 5}$

$\sf{\Rightarrow 10 - 5 = 5a}$

$\sf{\Rightarrow 5 = 5a}$

$\sf{\Rightarrow \dfrac{5}{5} = a}$

$\sf{\Rightarrow \dfrac{\not{5}}{\not{5}} = a}$

$\purple{\sf{\Rightarrow 1 ms^{-2} = a}}$

Hence , the Acceleration produced is 1 m/s².

Distance Covered :

• Initial velocity (u) = 5 m/s
• Final velocity (v) = 10m/s
• Acceleration (a) = 1/ms²

Let the distance covered be x m

Using the Third Equation of Motion , and Substituting the values in it ,we get :

$\purple{\sf{v^{2} = u^{2} + 2aS}}$

$\sf{\Rightarrow 10^{2} = 5^{2} + 2 \times 1 \times x}$

$\sf{\Rightarrow 10^{2} - 5^{2} = 2x}$

$\sf{\Rightarrow \dfrac{10^{2} - 5^{2}}{2} = x}$

$\sf{\Rightarrow \dfrac{100 - 25}{2} = x}$

$\sf{\Rightarrow \dfrac{75}{2} = x}$

$\purple{\sf{\Rightarrow 37.5 m = x}}$

Hence ,the distance covered by the car is 37.5 m.

» Additional information :

• Second Equation of Motion : s = ut + ½at².

• First Equation of Motion (u = 0) = v = at

• Height = ½gt²

• Height max. = u²/2g