Question

you are looking at new cars and a foolish salesman lets you take one for a test drive while driving you accelerate from rest at 2.2m/s2 for 5.2s. you then maintain this speed for 3.5s, then accelerate at 3.8m/s2 for 4.5s. how far have you travel

Answer 1

The total distance traveled by the car, considering the phases of acceleration and maintaining speed, was found to be 153.908 m.

Step-by-step explanation:

To calculate how far the car has traveled, we need to consider each phase of the trip separately and sum the distances covered in each phase.

Phase 1: Acceleration from rest

The car accelerates from rest at 2.2 m/s2 for 5.2 seconds. Using the equation of motion s = ut + (1/2)at2, where u is the initial speed (0 m/s since the car is at rest), a is the acceleration, and t is the time, we find the distance traveled in this phase as:

s1 = 0 * 5.2 + (1/2) * 2.2 * 5.22 = 29.808 m.

Phase 2: Maintaining Speed

The car then maintains this speed for 3.5 seconds. To find the distance covered in this phase, we calculate it by multiplying the speed at the end of the acceleration phase by the time maintained:

First, find the final speed after Phase 1 using v = u + at: v = 0 + 2.2 * 5.2 = 11.44 m/s.

The distance covered in Phase 2 is:

s2 = speed * time = 11.44 * 3.5 = 40.04 m.

Phase 3: Additional Acceleration

The car accelerates again, now at 3.8 m/s2 for 4.5 seconds. Similarly, we calculate the distance covered during this acceleration using the initial speed at the beginning of this phase (11.44 m/s) and the acceleration:

s3 = ut + (1/2)at2 = 11.44 * 4.5 + (1/2) * 3.8 * 4.52 = 84.06 m.

The total distance traveled is the sum of all three distances:

Total distance = s1 + s2 + s3 = 29.808 m + 40.04 m + 84.06 m = 153.908 m.