Question

two cars travel on a straight road from the point. A to point B both cars accelerate to their maximum speed and then continue at the speed for the rest of the distance. car 1 accelerates from rest to 20m/s over the 30s. It reaches point B in the 30s Wich car uses a larger acceleration to reach its maximum speed and which car has a larger average speed

Answer 1

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

Step-by-step 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