A car starts from rest and travels for 5.0 s with a uniform acceleration of +2.5 m/s2. The driver then applies the brakes, causing a uniform negative acceleration of -3.0 m/s2. If the brakes are applied for 2.0 s, how fast is the car going at the end of the braking period, and how far has it gone?
Well, well, looks like this car is going on a roller coaster ride of accelerations! Let's break it down, shall we?
First, we need to find out how fast the car is going after the initial 5 seconds of acceleration. We can use the good old formula: v = u + at, where v is the final velocity, u is the initial velocity (which is 0 since the car starts from rest), a is the acceleration (2.5 m/s^2 in this case), and t is the time (5 seconds). Plugging in the values, we get:
v = 0 + 2.5 * 5 = 12.5 m/s
So, after the initial acceleration, the car is zooming along at 12.5 m/s.
Now, the driver decides to put on the brakes. For the negative acceleration, we can still use the same formula, but with the negative value of acceleration (-3.0 m/s^2) and the braking time (2 seconds) this time:
v = 12.5 + (-3.0) * 2 = 12.5 - 6 = 6.5 m/s
So, after applying the brakes for 2 seconds, the car slows down to a speed of 6.5 m/s.
Now, let's figure out how far the car has gone during the whole process. To do that, we can use the formula: s = ut + 0.5at^2. Since the initial velocity is 0, the formula simplifies to s = 0.5at^2.
For the first 5 seconds of acceleration:
s1 = 0.5 * 2.5 * (5^2) = 31.25 m
For the 2 seconds of braking:
s2 = 0.5 * (-3.0) * (2^2) = -6.0 m (negative because it's going backward)
So, the total distance covered by the car is:
s_total = s1 + s2 = 31.25 - 6.0 = 25.25 m
So, at the end of the braking period, the car is going at a speed of 6.5 m/s and it has covered a total distance of 25.25 meters.
To solve this problem step-by-step, let's break it down into two parts: the acceleration phase and the braking phase.
Acceleration Phase:
1. First, let's find the final velocity after the acceleration phase. We can use the formula:
final velocity = initial velocity + (acceleration * time)
Since the car starts from rest (initial velocity = 0 m/s), the formula becomes:
final velocity = 0 + (2.5 m/s^2 * 5.0 s) = 12.5 m/s
2. Next, let's find the distance traveled during the acceleration phase. We can use the formula:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Again, since the car starts from rest (initial velocity = 0 m/s), the formula becomes:
distance = (0 * 5.0 s) + (0.5 * 2.5 m/s^2 * (5.0 s)^2) = 31.25 m
Braking Phase:
1. The car is now traveling at 12.5 m/s, and the driver applies the brakes resulting in a negative acceleration of -3.0 m/s^2.
2. Let's find the final velocity after the braking phase using the same formula as before:
final velocity = 12.5 m/s + (-3.0 m/s^2 * 2.0 s) = 6.5 m/s
3. Finally, let's find the distance traveled during the braking phase using the same formula as before:
distance = (12.5 m/s * 2.0 s) + (0.5 * -3.0 m/s^2 * (2.0 s)^2) = 25.0 m
Therefore, at the end of the braking period, the car is going at a speed of 6.5 m/s, and it has traveled a total distance of 56.25 m.
To determine the final velocity of the car after the braking period and the distance it has traveled, we can break down the problem into two parts: the initial acceleration period and the braking period.
1. Initial acceleration period:
We can use the equation of motion to calculate the velocity at the end of this period, given that the car starts from rest:
v = u + at
u = initial velocity = 0 m/s (starting from rest)
a = acceleration = +2.5 m/s^2 (positive because the car is accelerating)
t = time = 5.0 s
Plugging in these values, we can calculate the velocity at the end of the initial acceleration period.
v = 0 + (2.5)(5.0) = 12.5 m/s
So, at the end of the initial acceleration period, the car has a velocity of 12.5 m/s.
To determine the distance covered during this period, we use the equation of motion:
s = ut + (1/2)at^2
Plugging in the values:
s = (0)(5.0) + (1/2)(2.5)(5.0)^2 = 31.25 m
Therefore, the car has traveled a distance of 31.25 meters during the initial acceleration period.
2. Braking period:
During the braking period, the car experiences uniform negative acceleration.
Using the same equation of motion, we can calculate the velocity at the end of the braking period.
u = 12.5 m/s (velocity at the end of the initial acceleration period)
a = acceleration = -3.0 m/s^2 (negative because the car is decelerating)
t = time = 2.0 s
Plugging in these values, we get:
v = 12.5 + (-3.0)(2.0) = 6.5 m/s
Therefore, at the end of the braking period, the car has a velocity of 6.5 m/s.
To calculate the distance traveled during the braking period, we use the same equation of motion:
s = ut + (1/2)at^2
Plugging in the values:
s = (12.5)(2.0) + (1/2)(-3.0)(2.0)^2 = 29.0 m
Hence, the car has traveled a distance of 29.0 meters during the braking period.
In summary, at the end of the braking period, the car is moving with a velocity of 6.5 m/s and has traveled a total of 29.0 meters.
V1 = a*t = 2.5 * 5 = 7.5 m/s.
a. V2 = V1 + a*t = 7.5 -(3 * 2) = 1.5 m/s. After the braking period.
b. d1 = 0.5a*t^2 = 0.5*2.5*5^2 = 31.25 m.
d2 = V1*t + 0.5a*t^2.
d2 = 7.5*2 - 0.5*3*2^2 = 9 m.
d = d1+d2 = 31.25 + 9 = 40.25 m.