A car starts from rest and travels for 5.0 s with a uniform acceleration of + 1.5 m/s2. The driver then applies the brakes, causing a uniform acceleration of -2.0 m/s2. If the brakes are applied for 3.0 s, calculate (a) how fast is the car going at the end of the braking period, and (b) how far has it gone?
Phase 1
Use "vf=a(t)+vi" to find final velocity of phase 1, which is also the initial velocity of phase 2. Then use "x=1/2(vi+vf)t" to find displacement of first phase, which will be added later to displacement of second phase.
a = 1.5 m/s^2
t = 5.0 s
vi = 0 m/s
vf = 1.5 (5.0) + 0 = 7.5 m/s
x = 1/2 (0 + 7.5) (5.0) = 18.75 m
Phase 2
Use the final velocity found in phase 1 as the initial velocity of this phase. "vf=a(t)+vi" will help to find the final velocity after the braking period. Then use the final velocity to calculate the displacement of the second phase. Add the displacements of the two phases together.
a = -2.0 m/s^2
t = 3.0 s
vi = 7.5 m/s
vf = -2 (3.0) + 7.5 = 1.5 m/s
x = 1/2 (7.5 + 1.5) (3.0) = 13.5 m
18.75 m + 13.5 m = 32.25
The car is going 1.5 m/s at the end of the braking period. The car has travelled 32.25 m from its start.
(a) How fast is the car going at the end of the braking period?
Well, I hope the car doesn't end up going faster in reverse than it did in the forward direction! But let's crunch some numbers and find out.
First, let's calculate the final velocity during the first 5.0 seconds of acceleration. We can use the formula:
vf = vi + at
where vf is the final velocity, vi is the initial velocity (which is 0 m/s since the car starts from rest), a is the acceleration (1.5 m/s^2 in this case), and t is the time (5.0 seconds). Plugging in the values:
vf = 0 + (1.5 m/s^2)(5.0 s)
vf = 7.5 m/s
So, the car is going 7.5 m/s after the first 5.0 seconds of acceleration.
Now, let's calculate the final velocity during the braking period. Again, we can use the same formula:
vf = vi + at
This time, the initial velocity (vi) will be the final velocity at the end of the first 5.0 seconds of acceleration (which is 7.5 m/s), the acceleration (a) will be the negative value of the braking acceleration (-2.0 m/s^2), and the time (t) will be the duration of the braking period (3.0 seconds). Plugging in the values:
vf = 7.5 m/s + (-2.0 m/s^2)(3.0 s)
vf = 7.5 m/s - 6.0 m/s
vf = 1.5 m/s
The car is going 1.5 m/s at the end of the braking period.
(b) How far has it gone?
To find the distance traveled during the first 5.0 seconds of acceleration, we can use the equation:
d = vit + 0.5at^2
where d is the distance, vi is the initial velocity (0 m/s), t is the time (5.0 seconds), and a is the acceleration (1.5 m/s^2). Plugging in the values:
d = (0 m/s)(5.0 s) + 0.5(1.5 m/s^2)(5.0 s)^2
d = 0 m + 0.5(1.5 m/s^2)(25 s^2)
d = 0 m + 0.5(37.5 m)
d = 0 m + 18.75 m
d = 18.75 m
So, the car has traveled 18.75 meters during the first 5.0 seconds of acceleration.
To find the distance traveled during the braking period, we can use the same equation:
d = vit + 0.5at^2
This time, the initial velocity (vi) will be the final velocity at the end of the first 5.0 seconds of acceleration (7.5 m/s), the acceleration (a) will be the negative value of the braking acceleration (-2.0 m/s^2), and the time (t) will be the duration of the braking period (3.0 seconds). Plugging in the values:
d = (7.5 m/s)(3.0 s) + 0.5(-2.0 m/s^2)(3.0 s)^2
d = 22.5 m + 0.5(-2.0 m/s^2)(9.0 s^2)
d = 22.5 m + 0.5(-18.0 m)
d = 22.5 m - 9.0 m
d = 13.5 m
So, the car has traveled an additional 13.5 meters during the braking period.
Therefore, the total distance traveled by the car is:
18.75 m + 13.5 m = 32.25 m
The car has gone 32.25 meters in total.
To solve this problem, we need to break it down into different steps.
Step 1: Calculate the velocity of the car after the initial acceleration.
Given:
Initial velocity, u = 0 m/s
Acceleration, a = +1.5 m/s^2
Time, t = 5.0 s
We can use the equation of motion:
v = u + at
Substituting the given values, we have:
v = 0 + (1.5)(5.0)
v = 7.5 m/s
So, at the end of the initial acceleration, the car's velocity is 7.5 m/s.
Step 2: Calculate the distance covered during the initial acceleration.
Again, using the equation of motion:
s = ut + (1/2)at^2
Substituting the given values:
s = 0(5.0) + (1/2)(1.5)(5.0)^2
s = 37.5 m
Therefore, the distance covered during the initial acceleration is 37.5 m.
Step 3: Calculate the velocity of the car after the braking period.
Given:
Initial velocity after the initial acceleration, u = 7.5 m/s
Acceleration during braking, a = -2.0 m/s^2
Time of braking, t = 3.0 s
Using the equation of motion:
v = u + at
Substituting the given values:
v = 7.5 + (-2.0)(3.0)
v = 1.5 m/s
So, the car's velocity at the end of the braking period is 1.5 m/s.
Step 4: Calculate the distance covered during the braking period.
Using the equation of motion:
s = ut + (1/2)at^2
Substituting the given values:
s = 7.5(3.0) + (1/2)(-2.0)(3.0)^2
s = 22.5 - 9.0
s = 13.5 m
Therefore, the distance covered during the braking period is 13.5 m.
In conclusion,
(a) The car is going at a speed of 1.5 m/s at the end of the braking period.
(b) The car has traveled a distance of 13.5 m.
To solve this problem, we can use the equations of motion to find the final velocity and the displacement of the car. The equations of motion for uniformly accelerated motion are:
1. v = u + at
2. s = ut + (1/2)at^2
3. v^2 = u^2 + 2as
where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time
- s is the displacement
(a) To find the final velocity of the car after the braking period:
1. Calculate the initial velocity before braking:
The car starts from rest, so the initial velocity (u) is 0 m/s.
2. Calculate the velocity after the acceleration during the first 5.0 s:
Using equation (1), v = u + at, we have:
v = 0 + (1.5 m/s^2)(5.0 s)
v = 7.5 m/s
3. Calculate the velocity after the deceleration during the next 3.0 s:
Using equation (1), v = u + at, we have:
v = 7.5 m/s + (-2.0 m/s^2)(3.0 s)
v = 7.5 m/s - 6.0 m/s
v = 1.5 m/s
Therefore, the car is going at a speed of 1.5 m/s at the end of the braking period (negative sign indicates opposite direction).
(b) To find the distance travelled by the car:
1. Calculate the displacement during the first 5.0 s:
Using equation (2), s = ut + (1/2)at^2, we have:
s = 0 + (1/2)(1.5 m/s^2)(5.0 s)^2
s = (1/2)(1.5 m/s^2)(25.0 s^2)
s = 18.75 m
2. Calculate the displacement during the next 3.0 s:
Using equation (2), s = ut + (1/2)at^2, we have:
s = 7.5 m/s(3.0 s) + (1/2)(-2.0 m/s^2)(3.0 s)^2
s = 22.5 m - 9.0 m
s = 13.5 m
Therefore, the car has traveled a total distance of 18.75 m + 13.5 m = 32.25 m.
v = Vi + a t
x = X1 + Vi t +(1/2) a t^2
phase1
Xi = 0
Vi = 0
when t = 5 with a = 1.5
v = 0 + 5 t = 5*1.5 = 7.5 m/s
x = 0 + 0 + (1/2) (1.5) (25) = 18.75 meters
Phase 2
Xi = 18.75
Vi = 7.5
a = -2
t = 3
v = 7.5 - 2* 3 = 1.5 m/s
x = 18.75 + 7.5 *3 -(1/2)(2)(9)
= 18.75 + 22.5 - 9
= 13.5