A car accelerates steadly from zero to 30m/s in a distance d & a time t, another car takes a time 3t to accelerate steadly from stationary to the same final velocity.what distance does the second car cover during the new the new acceleration?
The second car has 1/3 of the acceleration of the first car.
In the same length of time, the second car travels 1/3 as far.
If they both travel to the same final velocity, the second car travels 3 times longer (in time), and goes (1/3)*3^2 or 3 times a far as the first car.
The distance travelled is 3d.
To find the distance covered by the second car during its acceleration phase, we can use the equation of motion relating distance, time, and acceleration.
Let's assume the acceleration of both cars to be "a" and the final velocity to be "v" (which is 30 m/s in this case).
For the first car, using the equation of motion:
v = u + at,
where u is the initial velocity (which is 0 m/s), we can rearrange the equation to find the acceleration:
a = (v - u) / t.
Substituting the given values:
a = (30 - 0) / t,
a = 30 / t.
Now, let's find the distance covered by the first car. We can use the equation of motion:
s = ut + (1/2)at^2,
where s is the distance covered, u is the initial velocity (0 m/s), a is the acceleration, and t is the time taken. Since the final velocity is 30 m/s, the distance covered by the first car is equal to "d".
Therefore, we have:
d = (0 * t) + (1/2)(30 / t)(t^2),
d = 15t.
Now, let's consider the second car. We know its acceleration is the same as the first car's final velocity divided by three times the time (3t).
Using the same equation of motion for the second car:
s = ut + (1/2)at^2,
where s is the distance covered, u is the initial velocity (0 m/s), a is the acceleration (30 / (3t)), and t is the time taken (3t).
Therefore, the distance covered by the second car is:
s = 0 * (3t) + (1/2)(30 / (3t))((3t)^2),
s = (1/2)(30)(3t),
s = 45t.
So, the second car covers a distance of 45t during its acceleration phase.