MATH SOLVE

9 months ago

Q:
# Two cars, Red and Blue, participate in a race. The Red car accelerates from 0 to 40 m/s in 5 seconds, while the Blue car accelerates from 0 to 35 m/s in 4 seconds. After the initial acceleration, both cars maintain their constant speed. Answer: a) Which car travels a greater distance after 10 seconds? b) What is the acceleration of both cars?

Accepted Solution

A:

Let's analyze the motion of both cars step by step:
a) Which car travels a greater distance after 10 seconds?
To find out which car travels a greater distance after 10 seconds, we'll calculate the distance traveled by each car during their initial acceleration and then during the constant speed phase.
For the Red car:
Initial acceleration: a = (final velocity - initial velocity) / time
a = (40 m/s - 0 m/s) / 5 s
a = 8 m/s²
Distance during initial acceleration:
d1 = (initial velocity * time) + (0.5 * acceleration * time²)
d1 = (0 m/s * 5 s) + (0.5 * 8 m/s² * (5 s)²)
d1 = 0 m + 0.5 * 8 m/s² * 25 s²
d1 = 100 m
Distance during constant speed:
d2 = (constant speed * time)
d2 = 40 m/s * 5 s
d2 = 200 m
Total distance traveled by the Red car after 10 seconds:
Total distance = d1 + d2 = 100 m + 200 m = 300 m
For the Blue car:
Initial acceleration: a = (final velocity - initial velocity) / time
a = (35 m/s - 0 m/s) / 4 s
a = 8.75 m/s²
Distance during initial acceleration:
d1 = (initial velocity * time) + (0.5 * acceleration * time²)
d1 = (0 m/s * 4 s) + (0.5 * 8.75 m/s² * (4 s)²)
d1 = 0 m + 0.5 * 8.75 m/s² * 16 s²
d1 = 70 m
Distance during constant speed:
d2 = (constant speed * time)
d2 = 35 m/s * 6 s
d2 = 210 m
Total distance traveled by the Blue car after 10 seconds:
Total distance = d1 + d2 = 70 m + 210 m = 280 m
So, after 10 seconds, the Red car travels a greater distance (300 meters) compared to the Blue car (280 meters).
b) What is the acceleration of both cars?
The acceleration of both cars during their initial acceleration phases is as follows:
Red car: 8 m/s²
Blue car: 8.75 m/s²