A car starts from rest and travels for 7.5 s
with a uniform acceleration of +1.9 m/s2
The driver then applies the brakes, causing a
uniform deceleration of 1.1 m/s2
If the brakes are applied for 2.9 s, how fast
is the car going at the end of the braking
period?
Answer in units of m/s
That’s wrong
To determine the final speed of the car after the braking period, we need to break down the problem into two parts: the acceleration phase and the deceleration phase.
First, let's calculate the distance covered during the acceleration phase. We can use the equation of motion:
\[ d = v_i \cdot t + \frac{1}{2} \cdot a \cdot t^2 \]
where:
- \( d \) is the distance covered,
- \( v_i \) is the initial velocity,
- \( a \) is the acceleration, and
- \( t \) is the time.
Since the car starts from rest, the initial velocity is 0 m/s and the time of acceleration is 7.5 s. Plugging in the values, we have:
\[ d = 0 \cdot 7.5 + \frac{1}{2} \cdot 1.9 \cdot (7.5)^2 \]
Simplifying the equation gives us:
\[ d = 0 + \frac{1}{2} \cdot 1.9 \cdot 56.25 \]
\[ d = 53.4375 \, \text{m} \]
Now, we can move on to the deceleration phase. Again, we use the equation of motion:
\[ v_f = v_i + a \cdot t \]
where:
- \( v_f \) is the final velocity,
- \( v_i \) is the initial velocity (which is the final velocity from the acceleration phase),
- \( a \) is the deceleration (negative since it opposes the motion), and
- \( t \) is the time.
We know that the deceleration is -1.1 m/s², the time is 2.9 s, and the initial velocity is the final velocity from the acceleration phase, which is given by the equation of motion:
\[ v_i = a \cdot t \]
Substituting the values, we have:
\[ v_i = 1.9 \cdot 7.5 \]
\[ v_i = 14.25 \, \text{m/s} \]
Now we can find the final velocity during deceleration using:
\[ v_f = v_i + a \cdot t \]
\[ v_f = 14.25 + (-1.1) \cdot 2.9 \]
Simplifying the equation gives us:
\[ v_f = 14.25 + (-3.19) \]
\[ v_f = 11.06 \, \text{m/s} \]
Therefore, at the end of the braking period, the car is going at a speed of 11.06 m/s.