A car starts from rest and travels for 5.9 s with a uniform acceleration of +1.2 m/s2. The driver then applies the brakes, causing a uniform acceleration of -1.7 m/s2. The breaks are applied for 1.70 s.
x0 = 0
v0 = 0
v1 = 1.2 * 5.9 = 7.08 m/s
x1 = (1/2)(1.2)(5.9^2) = 20.9 m
v3 = 7.08 - 1.7(1.7) = 4.19 m/s
x3 = 20.9 + 7.08 (1.7) -.5(1.7)(1.7)^2
= 30.5 m
To solve this problem, we can break it down into two parts: the car's motion before the brakes are applied and the car's motion while the brakes are applied.
Let's start with the car's motion before the brakes are applied.
Given:
Initial velocity (u) = 0 (as the car starts from rest)
Time (t1) = 5.9 s
Acceleration (a1) = +1.2 m/s²
We need to find the final velocity (v1) of the car after 5.9 seconds.
We can use the equation of motion:
v1 = u + a1 * t1
Substituting the values, we get:
v1 = 0 + 1.2 * 5.9
v1 = 7.08 m/s
So, the final velocity of the car before applying the brakes is 7.08 m/s.
Now, let's move on to the car's motion while the brakes are applied.
Given:
Initial velocity (u) = 7.08 m/s (as calculated earlier)
Time (t2) = 1.7 s
Acceleration (a2) = -1.7 m/s²
We need to find the final velocity (v2) of the car after 1.7 seconds of applying the brakes.
Using the same equation of motion:
v2 = u + a2 * t2
Substituting the values, we get:
v2 = 7.08 + (-1.7) * 1.7
v2 = 7.08 - 2.89
v2 = 4.19 m/s
So, the final velocity of the car after applying the brakes for 1.7 seconds is 4.19 m/s.