A car starts from rest and travels for 5.10 s with a uniform acceleration of +1.40 m/s2. The driver then applies the brakes, causing a uniform acceleration of -2.01 m/s2. If the brakes are applied for 2.83 s, how fast is the car going at the end of the braking period?
Part A.) 1.45m/s
Part B.) 30.4 m
To find the final velocity of the car at the end of the braking period, we can use the equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
1. First, let's find the initial velocity during the acceleration phase.
Given:
u1 = 0 m/s (The car starts from rest)
a1 = +1.40 m/s^2
t1 = 5.10 s
Using the equation: v1 = u1 + a1t1
v1 = 0 m/s + (1.40 m/s^2)(5.10 s)
v1 = 7.14 m/s
So, the initial velocity at the end of the acceleration phase is 7.14 m/s.
2. Next, let's find the final velocity during the braking phase.
Given:
u2 = 7.14 m/s (The car's velocity at the end of the acceleration phase)
a2 = -2.01 m/s^2
t2 = 2.83 s
Using the equation: v2 = u2 + a2t2
v2 = 7.14 m/s + (-2.01 m/s^2)(2.83 s)
v2 = 1.36 m/s
So, the final velocity at the end of the braking period is 1.36 m/s.
To find the final velocity of the car after the braking period, we need to break the problem down into two parts: the initial acceleration phase and the braking phase.
1. Initial acceleration phase:
During this phase, the car accelerates uniformly with an acceleration of +1.40 m/s^2 for a duration of 5.10 s from rest (initial velocity, u = 0).
Using the kinematic equation:
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 of the car at the end of the initial acceleration phase.
Substituting the known values:
v1 = 0 + (1.40 m/s^2)(5.10 s)
v1 = 7.14 m/s (rounded to two decimal places)
2. Braking phase:
During this phase, the car decelerates uniformly with an acceleration of -2.01 m/s^2 for a duration of 2.83 s. We can use the same kinematic equation to find the final velocity at the end of the braking phase.
Substituting the known values:
v2 = v1 + a(t)
where v1 is the final velocity at the end of the initial acceleration phase, a is the deceleration, and t is the time.
v2 = 7.14 m/s + (-2.01 m/s^2)(2.83 s)
v2 = 1.27 m/s (rounded to two decimal places)
Therefore, the car is going at a speed of 1.27 m/s at the end of the braking period.
part 1: v = 0+at = 1.40 * 5.10 = 7.14 m/s
part 2: v = 7.14 + at = 7.14 - 2.01*2.83 = 1.45 m/s