A car starts from rest and travels for 10.0 s with a uniform acceleration of +2.2 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?
m/s
(b) How far has the car gone?
m
To find the answers to these questions, we can use the equations of motion. Let's break down the problem step by step.
(a) How fast is the car going at the end of the braking period?
To find the final velocity at the end of the braking period, we need to calculate the change in velocity during the acceleration phase and the braking phase separately.
First, let's find the final velocity during the acceleration phase.
Given: Initial velocity (u) = 0 m/s, acceleration (a) = +2.2 m/s^2, time (t) = 10.0 s
Using the equation: v = u + at, where v is the final velocity,
v = 0 + 2.2 * 10.0
v = 22.0 m/s
Now, let's find the final velocity during the braking phase.
Given: Initial velocity (u) = 22.0 m/s (from the acceleration phase), acceleration (a) = -2.5 m/s^2, time (t) = 2.0 s
Using the same equation: v = u + at,
v = 22.0 + (-2.5) * 2.0
v = 22.0 - 5.0
v = 17.0 m/s
Therefore, the car is going at a speed of 17.0 m/s at the end of the braking period.
(b) How far has the car gone?
To find the distance traveled by the car, we need to calculate the distance covered during the acceleration phase and the braking phase separately.
First, let's find the distance covered during the acceleration phase.
Using the equation: s = ut + (1/2)at^2, where s is the distance covered,
s = 0 * 10.0 + (1/2) * 2.2 * (10.0)^2
s = 0 + 0.5 * 2.2 * 100.0
s = 110.0 m
Now, let's find the distance covered during the braking phase.
Using the same equation: s = ut + (1/2)at^2,
s = 22.0 * 2.0 + (1/2) * (-2.5) * (2.0)^2
s = 44.0 - 1.25
s = 42.75 m
Therefore, the car has traveled a total distance of 110.0 m + 42.75 m = 152.75 m.