1) two cars start at the same location and accelerate for the same amount of time. if one car has twice the acceleration of the other, how will the final speeds of the cars compare? by how much farther will the faster car outdistance the slower car?
Hmmm.... What is the MPH or KM of the slower car? If you do not know do not worry, I have the solution.
the question is like this! It doesnt have any units.
DA ANSWER IS OKAYYYYYY?
Y NO?
B=1/2A
THESE EQUATIOON IS SIMPLE COMMON SENSING!
DONT BE STOOPID!
THEN JUST PULGING WHAT YOUR KNOW?
DA ANSWER IS OKAYYYYYY?
Y NO?
THESE EQUATIOON IS SIMPLE COMMON SENSING!
DONT BE STOOPID!
THEN JUST PULGING WHAT YOUR KNOW?
The car with faster acceleration will be twice as far at all times
The car with faster acceleration will be twice as far at all times
To compare the final speeds of the cars, we can use the following equation of motion:
v = u + at
Where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time taken.
Since both cars start at the same location, their initial velocities (u) are zero. Let's assume the slower car has an acceleration of a1, and the faster car has an acceleration of 2a1 (twice the acceleration of the slower car).
For the slower car:
v1 = 0 + a1t
v1 = a1t
For the faster car:
v2 = 0 + 2a1t
v2 = 2a1t
From the equations above, we can see that the final speed of the faster car (v2) will be twice as much as the final speed of the slower car (v1).
To determine how much farther the faster car will outdistance the slower car, we need to calculate the distances covered by each car. The distance covered can be found using the following equation of motion:
s = ut + 0.5at²
Since both cars start from the same location, their initial distances (s) are zero. Let's find the distance covered by each car.
For the slower car:
s1 = 0.5a1t²
For the faster car:
s2 = 0.5(2a1)t²
s2 = 2(0.5a1t²)
s2 = a1t²
From the equations above, we can see that the faster car will outdistance the slower car by a factor of t².