A car starts from rest and travels for 5.0 s with a uniform acceleration of +1.9 m/s2. The driver then applies the brakes, causing a uniform acceleration of -3.0 m/s2. If the brakes are applied for 3.0 s, determine each of the following.
Well, let me calculate it for you, but be warned, my calculations are more like "calclownations"!
First, let's figure out the car's velocity after the initial acceleration phase. We can use the equation:
v = u + at,
where v is the final velocity, u is the initial velocity (which is zero in this case), a is the acceleration, and t is the time.
So, after 5.0 seconds of acceleration at +1.9 m/s², we have:
v₁ = 0 + (1.9 m/s²)(5.0 s).
v₁ = 9.5 m/s.
Now, let's figure out how far the car traveled during the acceleration phase using the equation:
s = ut + (1/2)at².
Since the initial velocity is zero, the equation simplifies to:
s = (1/2)at².
Plugging in the values, we have:
s₁ = (1/2)(1.9 m/s²)(5.0 s)².
s₁ = 23.8 m.
Now, let's determine the velocity after the phase of deceleration. Using the same equation as before:
v = u + at,
and substituting the values, we get:
v₂ = v₁ + (-3.0 m/s²)(3.0 s).
Calculating:
v₂ = 9.5 m/s + (-9.0 m/s).
v₂ = 0.5 m/s.
Finally, let's determine how far the car travels during the deceleration phase using the equation:
s = ut + (1/2)at².
Since the initial velocity is v₁, and the equation becomes:
s = v₁t + (1/2)at².
Substituting the values, we have:
s₂ = (9.5 m/s)(3.0 s) + (1/2)(-3.0 m/s²)(3.0 s)².
Calculating:
s₂ = 28.5 m + (-13.5 m).
s₂ = 15.0 m.
So, after all that calclownation, here are the results:
The velocity of the car after the initial acceleration phase is 9.5 m/s.
The distance traveled during the acceleration phase is 23.8 m.
The velocity of the car after the deceleration phase is 0.5 m/s.
The distance traveled during the deceleration phase is 15.0 m.
Hope that helps, and I hope you enjoyed my calclownations!
To determine each of the following, we can use the equations of motion and the information given:
1. The final velocity of the car before the brakes are applied:
Using the 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, we can find the final velocity before the brakes are applied.
Given:
- Initial velocity (u) = 0 m/s (the car starts from rest)
- Acceleration (a) = +1.9 m/s^2
- Time (t) = 5.0 s
Using the equation, v = u + at:
v = 0 + (1.9)(5.0)
v = 9.5 m/s
Therefore, the final velocity of the car before the brakes are applied is 9.5 m/s.
2. The distance traveled by the car before the brakes are applied:
Using the equation of motion s = ut + 1/2at^2, where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time, we can find the distance traveled before the brakes are applied.
Given:
- Initial velocity (u) = 0 m/s (the car starts from rest)
- Acceleration (a) = +1.9 m/s^2
- Time (t) = 5.0 s
Using the equation, s = ut + 1/2at^2:
s = (0)(5.0) + 1/2(1.9)(5.0)^2
s = 0 + 1/2(1.9)(25)
s = 23.75 m
Therefore, the distance traveled by the car before the brakes are applied is 23.75 m.
3. The final velocity of the car after the brakes are applied:
Using the 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, we can find the final velocity after the brakes are applied.
Given:
- Initial velocity (u) = 9.5 m/s (the final velocity before the brakes are applied)
- Acceleration (a) = -3.0 m/s^2
- Time (t) = 3.0 s
Using the equation, v = u + at:
v = 9.5 + (-3.0)(3.0)
v = 9.5 - 9.0
v = 0.5 m/s
Therefore, the final velocity of the car after the brakes are applied is 0.5 m/s.
4. The distance traveled by the car after the brakes are applied:
Using the equation of motion s = ut + 1/2at^2, where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time, we can find the distance traveled after the brakes are applied.
Given:
- Initial velocity (u) = 9.5 m/s (the final velocity before the brakes are applied)
- Acceleration (a) = -3.0 m/s^2
- Time (t) = 3.0 s
Using the equation, s = ut + 1/2at^2:
s = (9.5)(3.0) + 1/2(-3.0)(3.0)^2
s = 28.5 + 1/2(-3.0)(9.0)
s = 28.5 + 1/2(-27.0)
s = 28.5 - 13.5
s = 15.0 m
Therefore, the distance traveled by the car after the brakes are applied is 15.0 m.
To solve this problem, we need to break it down into two parts: the motion during the acceleration phase and the motion during the deceleration phase.
1. Motion during the acceleration phase:
Using the kinematic equation:
v = u + at
We know:
Initial velocity (u) = 0 m/s (the car starts from rest)
Acceleration (a) = +1.9 m/s^2
Time (t) = 5.0 s
Calculating the final velocity (v) during the acceleration phase:
v = u + at
v = 0 + (1.9)(5.0)
v = 9.5 m/s
Next, we can use another kinematic equation to calculate the distance covered (s) during the acceleration phase:
s = ut + (1/2)at^2
Plugging in the values:
s = (0)(5.0) + (1/2)(1.9)(5.0)^2
s = 0 + (0.5)(1.9)(25)
s = 0 + (0.5)(47.5)
s = 23.75 m
2. Motion during the deceleration phase:
During this phase, the acceleration is -3.0 m/s^2.
Calculating the final velocity (v) during the deceleration phase:
We can again use the kinematic equation:
v = u + at
We know:
Initial velocity (u) = 9.5 m/s (from the end of the acceleration phase)
Acceleration (a) = -3.0 m/s^2
Time (t) = 3.0 s
Plugging in the values:
v = u + at
v = 9.5 + (-3.0)(3.0)
v = 9.5 - 9.0
v = 0.5 m/s
Calculating the distance covered (s) during the deceleration phase:
Using the kinematic equation:
s = ut + (1/2)at^2
Plugging in the values:
s = (9.5)(3.0) + (1/2)(-3.0)(3.0)^2
s = 28.5 + (-4.5)(9)
s = 28.5 - 40.5
s = -12.0 m (Note: The negative sign indicates that the car moves in the opposite direction during deceleration phase)
Therefore, the answers to the questions are:
a) The final velocity after the acceleration phase is 9.5 m/s.
b) The distance covered during the acceleration phase is 23.75 m.
c) The final velocity after the deceleration phase is 0.5 m/s.
d) The distance covered during the deceleration phase is -12.0 m.