Please 🙏 break down number 3 question it's too long

Okay, let's break down question 3.

Question 3: A car starts from rest and accelerates uniformly at 2 m/s² for 10 seconds. It then travels at a constant velocity for 30 seconds. Finally, it decelerates uniformly to rest in 5 seconds. Calculate:

a) The maximum velocity reached by the car.

b) The total distance traveled by the car.

Here's the breakdown:

Part a) The maximum velocity reached by the car.

• Principle: We can use the first equation of motion for uniformly accelerated motion: v = u + at, where:

  • v = final velocity
  • u = initial velocity
  • a = acceleration
  • t = time

• Step 1: Identify the values for the first phase of motion (acceleration).

  • The car starts from rest, so the initial velocity (u) = 0 m/s.
  • The acceleration (a) = 2 m/s².
  • The time (t) for this phase = 10 seconds.

• Step 2: Substitute the values into the equation v = u + at.

  • v = 0 m/s + (2 m/s² × 10 s)

• Step 3: Calculate the final velocity.

  • v = 0 m/s + 20 m/s
  • v = 20 m/s

• Answer: The maximum velocity reached by the car is 20 m/s. This velocity is reached at the end of the first 10 seconds of acceleration.

Part b) The total distance traveled by the car.

This part requires calculating the distance for each of the three phases of motion and then summing them up.

• Phase 1: Acceleration (first 10 seconds)

  • Principle: We can use the equation of motion: s = ut + (1/2)at², where:
    • s = distance
    • u = initial velocity
    • t = time
    • a = acceleration
  • Step 1: Substitute values for Phase 1.
    • u = 0 m/s
    • t = 10 s
    • a = 2 m/s²
  • Step 2: Calculate distance for Phase 1 (s₁).
    • s₁ = (0 m/s × 10 s) + (1/2) × (2 m/s²) × (10 s)²
    • s₁ = 0 + (1 m/s²) × (100 s²)
    • s₁ = 100 m

• Phase 2: Constant Velocity (next 30 seconds)

  • Principle: For constant velocity, distance = velocity × time (s = vt).
  • Step 3: Identify values for Phase 2.
    • The velocity (v) is the maximum velocity reached in Phase 1, which is 20 m/s.
    • The time (t) for this phase = 30 s.
  • Step 4: Calculate distance for Phase 2 (s₂).
    • s₂ = 20 m/s × 30 s
    • s₂ = 600 m

• Phase 3: Deceleration (last 5 seconds)

  • Principle: We can use the equation of motion: v² = u² + 2as, or s = ut + (1/2)at². To use the second one, we first need to find the acceleration during deceleration.
  • Step 5: Calculate the acceleration (a) during deceleration.
    • Final velocity (v) = 0 m/s (comes to rest).
    • Initial velocity (u) = 20 m/s (velocity at the start of this phase).
    • Time (t) = 5 s.
    • Using v = u + at: 0 m/s = 20 m/s + a × 5 s
    • -20 m/s = a × 5 s
    • a = (-20 m/s) / 5 s = -4 m/s² (The negative sign indicates deceleration).
  • Step 6: Calculate distance for Phase 3 (s₃) using s = ut + (1/2)at².
    • u = 20 m/s
    • t = 5 s
    • a = -4 m/s²
    • s₃ = (20 m/s × 5 s) + (1/2) × (-4 m/s²) × (5 s)²
    • s₃ = 100 m + (-2 m/s²) × (25 s²)
    • s₃ = 100 m - 50 m
    • s₃ = 50 m

• Step 7: Calculate the total distance (S_total).

  • S_total = s₁ + s₂ + s₃
  • S_total = 100 m + 600 m + 50 m
  • S_total = 750 m

• Answer: The total distance traveled by the car is 750 m.