A car starts from rest and travels for 7.0 s with a uniform acceleration of +3.0 m/s2. The driver then applies the brakes, causing a uniform acceleration of −2.5 m/s2. If the brakes are applied for 2.0 s, determine each of the following.
(a) How fast is the car going at the end of the braking period?
I know the answer for this one is 16 m/s
(b) How far has the car gone?
How do you do this one?
To determine how far the car has gone during the braking period, we can use the equations of motion. Let's break down the problem into three parts: the car's acceleration phase, the constant velocity phase, and the braking phase.
1. Acceleration phase:
Given:
Initial velocity (u) = 0 m/s
Acceleration (a) = 3.0 m/s^2
Time (t) = 7.0 s
We can use the equation v = u + at to find the final velocity (v) after the acceleration phase:
v = 0 + (3.0 m/s^2)(7.0 s)
v = 21.0 m/s
2. Constant velocity phase:
During this phase, the car maintains a constant velocity, so there is no change in speed. Let's refer to the final velocity from the acceleration phase as the initial velocity for this phase.
Given:
Initial velocity (u) = 21.0 m/s
Time (t) = 2.0 s
Since there is no acceleration, the final velocity (v) will be the same as the initial velocity:
v = 21.0 m/s
To find the distance (S) traveled during this phase, we can use the equation S = ut, where u is the initial velocity and t is the time:
S = (21.0 m/s)(2.0 s)
S = 42.0 m
3. Braking phase:
Given:
Initial velocity (u) = 21.0 m/s
Acceleration (a) = -2.5 m/s^2
Time (t) = 2.0 s
Using the equation v = u + at, we can find the final velocity (v) after the braking phase:
v = 21.0 m/s + (-2.5 m/s^2)(2.0 s)
v = 21.0 m/s + (-5.0 m/s)
v = 16.0 m/s
Therefore, at the end of the braking period, the car is going at a speed of 16.0 m/s.
To find the total distance traveled by the car, we need to sum up the distances traveled during each phase. The distance traveled during the acceleration phase can be found using the equation S = ut + (1/2)at^2:
S1 = (0 m/s)(7.0 s) + (1/2)(3.0 m/s^2)(7.0 s)^2
S1 = 0 m + (1/2)(3.0 m/s^2)(49.0 s^2)
S1 = 72.75 m
Therefore, the total distance traveled by the car is the sum of S1, S2, and S3:
Total distance = S1 + S2 + S3
Total distance = 72.75 m + 42.0 m
Total distance = 114.75 m
Hence, the car has traveled a distance of 114.75 m.
The same way as the last one. At least you do not have centimeters
first phase, acceleration
u1 = 0 + a t1
x1 = 0 + 0 t1 + .5 a1 t1^2
second phase braking
u2 = u1 - 2.5(2)
x2 = x1 + u1(2) - .5 (2.5)(2)^2