A car is moving with a speed of 22 m/s. The driver then brakes, and the car comes to a halt after 6.5 s. What is the distance covered by the car after the driver brakes, until it comes to a stop?
a. 36 m
b. 72 m
c. 140 m
d. 219 m
acceleration --- (-a)
velocity = -at + c
when t = 0, v = 22
22= 0+c ---> c = 0
v = -at + 22
when t = 6.5, v = 0
0 = -6.5a + 22
a = 22/6.5 = 44/13
distance = d = (-1/2)a t^2 + 22t + k
d = (-22/13)t^2 + 22t + k
when t = 0, d = 0, so k = 0
when t = 0 , d = 0
when t = 6.5, d = (-22/13)(6.5)^2 + 22(6.5)
= 71.5 m
looks like they rounded it off to 72 m
To find the distance covered by the car after the driver brakes, until it comes to a stop, we need to use the equation for distance covered during uniform acceleration:
\[d = ut + \frac{1}{2}at^2\]
where:
- \(d\) is the distance covered,
- \(u\) is the initial velocity,
- \(t\) is the time taken, and
- \(a\) is the acceleration.
In this case, the car is initially moving with a speed of 22 m/s. When the driver brakes, the car decelerates, so the acceleration (a) will be negative.
Given:
\(u = 22\) m/s (initial velocity)
\(t = 6.5\) s (time taken)
\(a = ?\)
\(d = ?\)
The final velocity (v) is 0 m/s since the car comes to a stop.
Using the equation of motion:
\[v = u + at\]
Substituting the values:
\[0 = 22 + a \times 6.5\]
Simplifying:
\[a \times 6.5 = -22\]
\[a = -\frac{22}{6.5}\]
Now that we have the value of \(a\), we can substitute it into the equation for distance:
\[d = ut + \frac{1}{2}at^2\]
Substituting the known values:
\[d = 22 \times 6.5 + \frac{1}{2} \times -\frac{22}{6.5} \times (6.5)^2\]
Simplifying:
\[d = 143 + -22 \times 6.5 \times 6.5\]
\[d = 143 - 1014.5\]
\[d = -871.5\]
Since distance cannot be negative, we discard the negative sign and take the magnitude:
\[d = 871.5\]
Therefore, the distance covered by the car after the driver brakes, until it comes to a stop, is approximately 871.5 meters. Since none of the provided answer choices match exactly, we can conclude that none of the options are correct for this particular scenario.
To calculate the distance covered by the car after the driver brakes, until it comes to a stop, we can use the equation:
distance = average speed × time
Before braking, the car is moving with a speed of 22 m/s. After the brakes are applied, the car's speed decreases until it comes to a halt.
The average speed is the sum of the initial speed (22 m/s) and the final speed (0 m/s), divided by 2:
average speed = (initial speed + final speed) / 2 = (22 m/s + 0 m/s) / 2 = 22 m/s / 2 = 11 m/s
The time taken for the car to come to a halt is given as 6.5 s.
Let's substitute these values into the equation:
distance = average speed × time = 11 m/s × 6.5 s = 71.5 m
Therefore, the distance covered by the car after the driver brakes, until it comes to a stop, is approximately 71.5 meters.
The closest option from the given choices is b. 72 m.