Question

A CAR STARTS FROM REST AND ACCELERATES AT A UNIFORM RATE OF 5M/S SQUARE FOR SOME TIME , THEN MOVES WITH CONSTANT SPEED FOR SOME TIME AND THEN RETARDS AT THE SAME UNIFORM RATE OF 5M/S SQUARE AND COMES TO REST. TOTAL TIME TAKEN FOR THE JOURNEY IS 25 SECS AND THE AVERAGE SPEED IS 20 M/S . HOW LONG THE CAR MOVES WITH CONSTANT SPEED ?​

Answer 1

Answer:

15 sec

Explanation:

Let say Car accelerates at 5 m/s²  for T₁  sec

V = U + aT

Velocity attained = 0 + 5T₁ = 5T₁

S = ut + (1/2)aT²

Distance Covered = 5T₁²/2

Let say constant Speed = T₂ sec

Distance Covered in T₂ = 5T₁T₂

T₃ for Retardation =  (0 - 5T₁)/(-5) = T₁

T₃ = T₁

Distance Covered = 5T₁² - 5T₁²/2 = 5T₁²/2

T₁ + T₂ + T₁ = 25  => 2T₁ + T₂ = 25  => T₁ = (25 - T₂)/2

5T₁²/2 + 5T₁T₂ + 5T₁²/2 = 5T₁² + 5T₁T₂  = Total Distance covered

Average speed = 20 m/s

20*25 = 500 m

5T₁² + 5T₁T₂  = 500

=> T₁² + T₁T₂ = 100

putting  T₁ = (25 - T₂)/2

=> ((25 - T₂)/2)² + ((25 - T₂)/2)T₂ = 100

=> (25 -  T₂)² + 2(25 - T₂)T₂ = 400

=> 625 + T₂² - 50T₂ + 50T₂ - 2 T₂² = 400

=> T₂² = 225

=>  T₂ = 15

THE CAR MOVES WITH CONSTANT SPEED for 15 sec