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
