A car starts from rest and travels for 5.0 s with a steady acceleration of magnitude 1.5 m/s2. the driver then applies the brakes for 3.0 seconds causing a steady deceleration of magnitude 2.0 m/s2. How fast is the car going at the end of the breaking period?
V1 = a*t = 1.5 * 5 = 7.5 m/s.
V2 = V1 + a*t = 7.5 - 2*3 = 1.5 m/s.
To find the final velocity of the car at the end of the braking period, we will need to use the equations of motion.
1. First, let's calculate the acceleration during the constant acceleration phase using the equation:
v = u + at
where:
v = final velocity
u = initial velocity (in this case, since the car starts from rest, u = 0)
a = acceleration
t = time
Given:
a = 1.5 m/s²
t = 5.0 s
Substituting the given values into the equation, we get:
v = 0 + (1.5 m/s²)(5.0 s)
v = 7.5 m/s
So, the velocity at the end of the constant acceleration phase is 7.5 m/s.
2. Next, let's calculate the displacement during the constant acceleration phase using the equation:
s = ut + (1/2)at²
where:
s = displacement
u = initial velocity
a = acceleration
t = time
Since the car starts from rest (u = 0), the equation simplifies to:
s = (1/2)at²
Plugging in the values, we find:
s = (1/2)(1.5 m/s²)(5.0 s)²
s = (1/2)(1.5 m/s²)(25.0 s²)
s = (18.75 m²/s²)
So, the displacement during the constant acceleration phase is 18.75 meters.
3. Now, let's calculate the deceleration phase. Since the car is slowing down, we need to consider the negative sign (-2.0 m/s²) to represent deceleration.
Using the equation:
v² = u² + 2as
where:
v = final velocity
u = initial velocity
a = acceleration (deceleration in this case)
s = displacement
Since we want to find the final velocity (v), we rearrange the equation:
v = √(u² + 2as)
Given:
u = 7.5 m/s (velocity at the end of the constant acceleration phase)
a = -2.0 m/s² (deceleration)
s = -18.75 m (negative displacement, as the car is moving in the opposite direction)
Substituting the given values into the equation, we get:
v = √((7.5 m/s)² + 2(-2.0 m/s²)(-18.75 m))
v = √(56.25 m²/s² + 75 m²)
v = √(131.25 m²/s²)
v ≈ 11.46 m/s
So, the car's speed at the end of the braking period is approximately 11.46 m/s.