Answer:
Part A
The acceleration of car 2, is larger than the acceleration of car 1
Part B
The average speed of car 2 is larger than the average speed of car 1
Explanation:
The question relates to kinematic motion
Using a similar question for the parameters, we have;
The time it takes car 1 to accelerate to v₁ = 20 m/s is t₁ = 30 s
The time it takes car 1 to reach point B, t₁₂ = 35 s
The time it takes car 2 to accelerate to v₂ = 20 m/s is t₂ = 20 s
The time it takes car 2 to reach point B is t₂₂ = 30 s
Part A
From the kinematic equation of motion, v = u + at, we have;
Acceleration, a = (v - u)/t
Where;
a = The acceleration
v = The final velocity
u = The initial velocity = 0 m/s for both cars as they start from rest
t = The time taken
The acceleration of car 1, a₁ = (v₁ - u₁)/t₁
∴ a₁ = (20 m/s - 0 m/s)/(30 s) = 2/3 m/s²
The acceleration of car 2, a₂ = (v₂ - u₂)/t₂
∴ a₂ = (20 m/s - 0 m/s)/(20 s) = 1 m/s²
The acceleration of car 2, a₂ = 1 m/s² is larger than the acceleration of car 1, a₁ = 2/3 m/s²
Part B
Average speed = Total distance/(Total time taken to reach the distance)
Let 'B' represent the distance to point in meters, we have;
The average speed of car 1 = B/(The time it takes car 1 to reach point B)
∴ The average speed of car 1 = B/35 m/s
The average speed of car 2 = B/(The time it takes car 2 to reach point B)
∴ The average speed of car 2 = B/30 m/s
Noting that the larger velocity is given by the smaller divisor to 'B', we have;
The average speed of car 2 = B/30 m/s is larger than the average speed of car 1 = B/35 m/s
