A car starts from rest and travels for 10 seconds with a constant acceleration of 3.0 m s^-2. The driver then applies the brakes causing a negative acceleration of -4.0 m s^-2. Assuming that the brakes are applied for 2.0 seconds, calculate,
a) the velocity of the car at the end of braking
b) the distance travelled by the car at the end of braking
a) Well, it seems like this car had a bit of a roller coaster ride! To find the velocity at the end of braking, we'll need to break it down into two parts. First, we'll calculate the velocity during the initial acceleration phase, and then we'll calculate the velocity during the braking phase.
During the initial acceleration phase, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Since the car starts from rest, the initial velocity (u) is 0.
So, during the initial acceleration phase, we have v = 0 + (3.0 m/s^2)(10 s) = 30 m/s.
Now, during the braking phase, we'll need to consider the negative acceleration. Again, we can use the same equation, v = u + at, but this time the acceleration (a) is negative.
Since the braking lasts for 2 seconds, we have v = 30 m/s + (-4.0 m/s^2)(2 s) = 22 m/s.
So, the velocity of the car at the end of braking is 22 m/s.
b) To find the distance traveled by the car at the end of braking, we'll need to calculate the distance covered during the initial acceleration phase and the distance covered during the braking phase.
During the initial acceleration phase, we can use the equation s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Since the car starts from rest, the initial velocity (u) is 0. So, during the initial acceleration phase, we have s = 0 + (1/2)(3.0 m/s^2)(10 s)^2 = 150 m.
Now, during the braking phase, we can again use the same equation, s = ut + (1/2)at^2. Since the acceleration (a) is negative, we'll use the value of -4.0 m/s^2.
With a braking time of 2 seconds, we have s = 30 m/s(2 s) + (1/2)(-4.0 m/s^2)(2 s)^2 = 60 m + (-8 m) = 52 m.
So, the total distance traveled by the car at the end of braking is 150 m + 52 m = 202 m.
And there you have it! The velocity of the car at the end of braking is 22 m/s, and the distance traveled by the car at the end of braking is 202 m. Shall I tell you a car joke to wrap things up?
To solve this problem, we can use the kinematic equations of motion. There are three important equations that we will use:
1) v = u + at, which relates initial velocity (u), final velocity (v), acceleration (a), and time (t).
2) s = ut + (1/2)at^2, which relates displacement (s), initial velocity (u), acceleration (a), and time (t).
3) v^2 = u^2 + 2as, which relates final velocity (v), initial velocity (u), acceleration (a), and displacement (s).
Let's solve part (a) first:
a) The velocity of the car at the end of braking.
Given:
Initial velocity (u) = 0 m/s
Acceleration (a) = 3.0 m/s^2
Time (t) = 10 seconds
Using the equation v = u + at, we can calculate the velocity at the end of the 10 seconds:
v = u + at
v = 0 + (3.0)(10)
v = 0 + 30
Therefore, the velocity of the car at the end of the 10 seconds is 30 m/s.
Now let's solve part (b):
b) The distance traveled by the car at the end of braking.
Given:
Initial velocity (u) = 0 m/s
Acceleration (a) = -4.0 m/s^2 (negative because it's deceleration due to braking)
Time (t) = 2 seconds
Since we are looking for distance traveled, we will use the equation s = ut + (1/2)at^2:
s = ut + (1/2)at^2
s = 0 + (1/2)(-4.0)(2^2)
s = 0 + (1/2)(-4.0)(4)
s = 0 + (-2.0)(4)
s = 0 - 8
s = -8
The negative sign indicates that the car has traveled 8 meters in the opposite direction.
Therefore, the distance traveled by the car at the end of braking is 8 meters in the opposite direction.
To solve this problem, we can break it down into three separate parts:
1. The car's motion during acceleration
2. The car's motion during braking
3. Combining the two motions to find the final velocity and distance traveled
Let's start with the car's motion during acceleration.
1. The car's motion during acceleration:
We are given that the car starts from rest and travels for 10 seconds with a constant acceleration of 3.0 m/s². We can use the kinematic equation to find the car's velocity at the end of this period:
v = u + at
Where:
v is the final velocity
u is the initial velocity (0 m/s, since the car starts from rest)
a is the acceleration (3.0 m/s²)
t is the time (10 seconds)
Plugging in the values, we get:
v = 0 + (3.0 m/s²) * (10 s)
v = 0 + 30 m/s
v = 30 m/s
So, the car's velocity at the end of the acceleration phase is 30 m/s.
2. The car's motion during braking:
Next, we need to calculate the car's motion during braking. We are given that the brakes are applied for 2.0 seconds with a negative acceleration of -4.0 m/s². Since the car was initially moving at 30 m/s, we can use the same kinematic equation to find the final velocity during braking:
v = u + at
Where:
v is the final velocity
u is the initial velocity (30 m/s)
a is the acceleration (-4.0 m/s²)
t is the time (2.0 seconds)
Plugging in the values, we get:
v = 30 + (-4.0 m/s²) * (2 s)
v = 30 - 8 m/s
v = 22 m/s
So, the car's velocity at the end of braking is 22 m/s.
3. Combining the two motions to find the final velocity and distance traveled:
To find the final velocity of the car, we need to consider both the acceleration and braking phases. Since acceleration and braking have opposite directions, we can simply subtract the magnitude of the braking velocity from the magnitude of the acceleration velocity:
Final velocity = |Acceleration velocity| - |Braking velocity|
Final velocity = |30 m/s| - |22 m/s|
Final velocity = 30 m/s - 22 m/s
Final velocity = 8 m/s
To find the distance traveled during the acceleration and braking phases, we can use the following equation:
s = ut + (1/2)at²
Where:
s is the distance traveled
u is the initial velocity
t is the time
a is the acceleration
First, let's find the distance traveled during the acceleration phase:
s1 = (1/2)(u + v)t1
s1 = (1/2)(0 + 30 m/s)(10 s)
s1 = (1/2)(30 m/s)(10 s)
s1 = 150 m
Next, let's find the distance traveled during the braking phase:
s2 = ut2 + (1/2)at2²
s2 = (22 m/s)(2 s) + (1/2)(-4.0 m/s²)(2 s)²
s2 = 44 m + (-4 m/s²)(4 s²)
s2 = 44 m - 32 m
s2 = 12 m
Finally, we can find the total distance traveled by adding the distances from both phases:
Total distance = s1 + s2
Total distance = 150 m + 12 m
Total distance = 162 m
So, the velocity of the car at the end of braking is 8 m/s, and the distance traveled by the car at the end of braking is 162 m.
v = Vo + at
s = Vo*t + 1/2 at^2
so,
v = 0 + 3(10) + 2(-4) = 22 m/s
s = (0*10 + 1/2 * 3 * 10^2) + (30*2 - 1/2 * 4 * 2^2) = 202 m