Question

3) The initial velocity of a car is 10 m/s. It accelerates uniformly at 2 m/s2 for 20 s and there
after retards uniformly at 4 m/s2. Find the total distance covered by the car before it comes
to rest. For how much time the car is in motion? ​

Answer 1

Given

• Initial Velocity = 10 m/s
• Acceleration₁ = 2 m/s
• Time= 20 sec
• Then it starts regarding at a rate of 4 m/s²

To Find

• Distance Covered before it goes to rest
• Time in which the car is in motion

Solution

● v = u + at [First Equation of Motion]

● v²-u² = 2as [Third Equation of Motion]

● s = ut + ½at² [Second Equation of Motion]

✭ Final Velocity after the acceleration :

→ v = u + at

→ v = 10 + 2 × 20

→ v = 10 + 40

→ v = 50 m/s

✭ Distance Covered during this accn :

→ v²-u² = 2as

→ 50²-10² = 2 × 2 × s

→ 2500 - 100 = 4s

→ 2400 = 4s

→ 2400/4 = s

→ Distance = 600 m

✭ Time taken for the car to stop :

→ v = u + at

→ 0 = 50 + (-4) × t

→ -50/-4 = t

→ Time = 12.5 sec

✭ Total Time :

→ Total Time = Time in acceleration + Time in retardation

→ Total Time = 20 + 12.5

→ Time in Motion = 32.5 sec

✭ Distance Covered during retardation :

→ s = ut + ½at²

→ s = 50 × 12.5 + ½ × 4 × 12.5²

→ s = 625 + 2 × 156.25

→ s = 625 + 312.5

→ s = 937.5 m

✭ Total Distance :

→ Total Distance = Distance during acceleration + Distance during deceleration

→ Total Distance = 600 + 937.5

→ Total Distance = 1537.5 m

Answer 2

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