A car starts from rest and travels for 6.3 s with
a uniform acceleration of 2.3 m/s2. The driver
then applies the brakes, causing a uniform
acceleration of −1.9 m/s2.
If the brakes are applied for 2.6 s, how fast
is the car going at the end of the braking
period?
vₒ=a1•t1 = 6.3•2.3 =14.49 m/s
v=vₒ-a2•t2 = 14.49 – 1.9-2.6 = 9.55 m/s
To solve this problem, we can divide it into two parts: the initial acceleration phase and the braking phase.
Step 1: Find the final velocity after the initial acceleration phase.
In this phase, the car starts from rest and travels for 6.3 seconds with a uniform acceleration of 2.3 m/s^2. We can use the kinematic equation:
v = u + at
where:
v = final velocity
u = initial velocity (which is 0 since the car starts from rest)
a = acceleration
t = time
Plugging in the values, we have:
v = 0 + (2.3 m/s^2)(6.3 s)
v = 14.49 m/s
Step 2: Find the final velocity after the braking phase.
In this phase, the brakes are applied for 2.6 seconds, with a uniform acceleration of -1.9 m/s^2. Again, we can use the kinematic equation:
v = u + at
where:
v = final velocity
u = initial velocity (which is 14.49 m/s, the final velocity from the initial acceleration phase)
a = acceleration
t = time
Plugging in the values, we have:
v = 14.49 m/s + (-1.9 m/s^2)(2.6 s)
v = 14.49 m/s - 4.94 m/s
v = 9.55 m/s
Therefore, at the end of the braking period, the car is going at a speed of 9.55 m/s.
To find the final velocity of the car at the end of the braking period, we need to break down the problem into two parts: the initial acceleration phase and the braking phase.
1. Initial acceleration phase:
During this phase, the car starts from rest and travels for 6.3 seconds with a uniform acceleration of 2.3 m/s^2. We can use the following kinematic equation to find the final velocity (V) at the end of this phase:
V = u + at
where:
V = final velocity
u = initial velocity (since the car starts from rest, u = 0)
a = acceleration (2.3 m/s^2)
t = time taken (6.3 seconds)
Plugging in the values, we get:
V = 0 + (2.3 m/s^2) * (6.3 s)
V = 0 + 14.49 m/s
V = 14.49 m/s
The car's velocity at the end of the initial acceleration phase is 14.49 m/s.
2. Braking phase:
During this phase, the driver applies the brakes, causing a uniform acceleration of -1.9 m/s^2. We can again use the same kinematic equation to find the final velocity at the end of this phase. However, we need to consider that the deceleration is negative (-1.9 m/s^2). The time taken for the braking phase is given as 2.6 seconds.
V = u + at
where:
V = final velocity (which is what we are trying to find)
u = initial velocity (which is the final velocity at the end of the initial acceleration phase, i.e., 14.49 m/s)
a = acceleration (which is the deceleration caused by the brakes, i.e., -1.9 m/s^2)
t = time taken (2.6 seconds)
Plugging in the values, we get:
V = 14.49 m/s + (-1.9 m/s^2) * (2.6 s)
V = 14.49 m/s - 4.94 m/s
V = 9.55 m/s
The car's velocity at the end of the braking period is 9.55 m/s.