Determine the time required for a car to travel 1 km along a road if the car starts from rest, reaches a maximum speed at some intermediate point, and then stops at the end of the road. The car can accelerate at 1.5 m/s2 and decelerate at 2 m/s2.

Total time taken by car to travel 1 km distance is 48.2 seconds

Explanation:

Given data-

Total distance- 1 km= 1000m

Car starts from rest- Hence the initial velocity (u)= 0 m/s

Then, car accelerates at 1.5 m/s ²

Let us suppose with these acceleration car reaches the max speed of V

Car then decelerates at rate of 2 m/s ²

Finally, car comes to rest

We need to consider the question in two parts

Part 1= when car starts from rest and reaches the value V in the time tₐ at the distance s₁.

Part 2= When car starts from V and finally stops at 1 km mark in the time tₙ in distance 1000-s₁.

For the part 1

We know the formula  

v= u + a*tₐ

where v= final velocity

u- initial velocity

a= acceleration

tₐ= time period

At the starting u= 0  

Hence the equation reduces to V=0+1.5tₐ

Or V= 1.5tₐ                             Eq 1

We also know that s= u*tₐ+ ¹/₂*a*tₐ²

Where s₁= distance covered  (other symbols same meaning)

Since u=0   (u*tₐ=0)

s₁ = ¹/₂*1.5*tₐ²

s₁= ½* 1.5*tₐ²                    Eq 2

Now considering Part 2

Here the case is deceleration hence the equation would change (symbols same)

v= V-a*tₙ     final velocity(v)=0 (car stops finally) & initial velocity (part 2)= V

V= a*tₙ

V=2*tₙ                                 Eq 3

Similarly  

1000-s₁= V* tₙ+¹/₂*(-2) *(tₙ)²

1000-s₁= V* tₙ-tₙ²                      Eq 4

Comparing Eq 1 and Eq 3

V= 1.5tₐ  and V=2tₙ                                                              

1.5tₐ = 2 tₙ

tₐ=1.33 tₙ  

Using the above value of tₐ in Eq 1

V= 1.5 tₐ and tₐ= 1.33 tₙ

V= 2tₙ

Similarly from Eq 2 and putting the value of tₙ

s₁= ½*1.5*tₐ²      

s₁=  1.33*(tₙ)²

Substituting the above values in equation 4

1000-s₁= V* tₙ-tₙ²                      

1000- 1.33(tₙ)²=2*(tₙ)*(tₙ)- tₙ²

1000=2.33 (tₙ)²

tₙ=      1000/2.333    

tₙ= 20.7 sec

Similarly putting the value of tₙ in tₐ= 1.33 tₙ

tₙ= 27.5 sec

Hence total time is tₐ +tₙ  

T= 20.7+27.5= 48.2 sec                

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