To solve this problem, let's break it down into three phases: acceleration, constant speed, and deceleration.
Phase 1: Acceleration The car starts from rest and accelerates at a rate of 1 m/s² for 10 seconds. During this phase, we can use the following equation of motion:
d = v₀t + (1/2)at²
where: d is the distance traveled, v₀ is the initial velocity (0 m/s in this case), t is the time, and a is the acceleration (1 m/s²).
Plugging in the values, we have:
d₁ = (0)(10) + (1/2)(1)(10)² = 0 + (1/2)(1)(100) = 0 + 50 = 50 meters
Phase 2: Constant Speed The car continues at a steady speed for 20 seconds, which means it covers a distance equal to the product of the speed and time:
d₂ = v₂t₂
where: d₂ is the distance traveled during this phase, v₂ is the constant speed, and t₂ is the time (20 seconds in this case).
Phase 3: Deceleration The car decelerates and comes to rest in 5 seconds. Using the equation of motion mentioned earlier, but with the final velocity v = 0 m/s, we have:
d₃ = (0)(5) + (1/2)(-a)(5)² = 0 + (1/2)(-1)(25) = 0 - 12.5 = -12.5 meters
Note that the distance is negative in this case since it represents the opposite direction of motion.
Now, we need to find the time taken to cover half the distance. Let's denote it as tₕ.
The total distance traveled can be calculated as the sum of the distances in each phase:
d = d₁ + d₂ + d₃ = 50 + d₂ - 12.5
Half the total distance is equal to:
d/2 = (50 + d₂ - 12.5)/2 = (37.5 + d₂)/2
Now, we can set up an equation to find tₕ:
(37.5 + d₂)/2 = v₂tₕ
Since the speed is constant during the second phase, v₂tₕ represents the distance covered in the constant speed phase. Solving for tₕ:
tₕ = (37.5 + d₂)/(2v₂)
To calculate tₕ, we need to know the value of v₂ (the constant speed during phase 2). However, the problem does not provide this information, so we cannot determine the exact value of tₕ without additional data.