A car is moving with a uniform velocity of 20 m/s for 20 seconds. Then brakes are applied and it comes to rest with uniform deceleration in 1 minute.
a. Calculate the total distance covered by the car using a mathematical equation?
b. Plot the graph to calculate this distance using the speed time graph? 

Ans. {1000 m}



Method-1 ( by using the equations of motion): 


Given:  

From the question, it is clear that the car has traveled 2 distances, which are following 

(1) Travelled with speed = v = 20 m s⁻¹ for t = 20 s.

(2) Finally moves with uniform deceleration  (retardation) and is stopped after (1 minute = 60 s)

a) Distance S₁, A car is moving with uniform velocity of 20 m/s for 20 seconds.



Given: 
 
Velocity of the car = v = 20 m s⁻¹ 

Time = t = 20 s


To Find:

Distance = S₁ = ?


Solution:  

By using the equation

S₁ = vave ï½˜ t

by putting values

S₁ = 20 m s⁻¹ + 20 s

S₁ = 400 m ----------- eqn (1)



b) Distance S2, when, it moves with uniform deceleration (Retardation) and is stopped after 1 minute.


Given:  

Initial Velocity of the car = vi = 20 m s⁻¹ (as the car was moving initially with this velocity and now going to stop)

Time taken by the car = t = 1 min = 60 s 

Final Velocity of the car = vf = 0 m s⁻¹ (as it comes to rest)


To Find:

Distance = S₂ = ?


Solution:  

According to the above-given data, the Distance S₂  can be calculated by using both the 2nd or 3rd equation of motion.

But we will first find the retardation: as the car is going to stop, So vf = 0 ms⁻¹, and the velocity of the car at which it is traveling now will be vi = 20 m s⁻¹ and time = 1min = 60 sec

so using 1st equation of motion

vf = vi + at

or 

a = `\frac{vf - vi}{t}`

by putting values

a  = `\frac{0 m s⁻¹ - 20 m s⁻¹}{60 s}`

a  = -`\frac{ 20 m s⁻¹}{60 s}`

by simplifying we get

a = - 0.333 m s⁻²

Now by using the 3rd equation of motion

2aS₂  =  vf ² - vi ²

By putting values

2 (-0.33333 m s⁻²)  S₂ =  (0 m s⁻¹) ² - (20 m s⁻¹) ²

(-0.667 m s⁻²)  S₂ =   - 400 m² s⁻²

or 

S₂ = 599.7 m = 600 m -------------eqn (2)

Total distance traveled by the car
= S  = S1  +  S2  = 400 m + 600 m= 1000 m   ------------------Ans.



Method 2 (with the help of a diagram)


Given:  

From the question, it is clear that the car has traveled 2 distances, which are following 

(1) Travelled with speed = v = 20 m s⁻¹ for t = 20 s

(2) Finally it moves with uniform deceleration  (retardation) and is stopped after (1 minute = 60 s)



To Find:

Total Distance Traveled by car = S=?





Solution:  

According to the given conditions in the numerical, we can solve this problem easily by drawing a velocity-time graph as below by using the scale as
Along X-Axis :    One big division = 20 s
Along y-Axis :    One big division = 10 m s⁻¹ 




The area of the trapezium (OABC) formed in the velocity-time graph is numerically equal to the total distance covered by the body. that is

Total Distance Travelled = area of the trapezium (OABC) of the graph

Total Distance Travelled = `\frac {1}{2}` ｘ {sum of the parallel sides} ｘ height of Trapezium

Total Distance Travelled = `\frac {1}{2}` ｘ {80 s + 20 s} ｘ 20 m s⁻¹ 

Total Distance Travelled = `\frac {1}{2}` ｘ {100 s} ｘ 20 ms⁻¹ 

Total Distance Travelled = 50 s ｘ 20 m s⁻¹ 

Total Distance Travelled = 1000 m




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