A car having an initial speed of 16 m/s is uniformly brought to rest in 4.0 s. How far does the car travel during this 4.0 s interval?
a=changevelocity/time
Vf^2=Vi^2 +2ad solve now for d.
32
To find the distance traveled by the car, we can use the formula:
\[ d = ut + \frac{1}{2}at^2 \]
where:
- \( d \) is the distance traveled
- \( u \) is the initial velocity
- \( t \) is the time taken
- \( a \) is the acceleration
In this case, the initial velocity \( u \) is 16 m/s, the time \( t \) is 4.0 s, and the car is coming to a rest, so the final velocity is 0 m/s.
Since the car is coming to a rest, the acceleration \( a \) can be calculated using the formula:
\[ v = u + at \]
where:
- \( v \) is the final velocity
In this case, the final velocity \( v \) is 0 m/s.
Rearranging the formula, we have:
\[ 0 = 16 + 4a \]
Solving this equation, we find that \( a = -4 \) m/s².
Now we can substitute the known values into the distance formula:
\[ d = 16(4) + \frac{1}{2}(-4)(4)^2 \]
Simplifying, we have:
\[ d = 64 - 32 = 32 \text{ m} \]
Therefore, the car travels a distance of 32 meters during the 4.0-second interval.
To find the distance traveled by the car during the 4.0 s interval, you can use the equation of motion for uniformly accelerated motion. The equation is:
s = ut + (1/2)at^2
Where:
s = distance traveled
u = initial velocity
t = time interval
a = acceleration
In this case, the initial velocity (u) is 16 m/s, the time interval (t) is 4.0 s, and the car comes to rest, so the final velocity is 0 m/s. Since the car is brought uniformly to rest, the acceleration (a) will be negative.
We can calculate the acceleration using another equation:
v = u + at
Rearranging the equation to solve for acceleration gives:
a = (v - u) / t
Substituting the values, we have:
a = (0 - 16) / 4.0
a = -4 m/s^2
Now, we can substitute the values of u, a, and t into the equation of motion:
s = (16 * 4.0) + (1/2)(-4)(4.0)^2
s = 64.0 - 32.0
s = 32.0 m
Therefore, the car travels a distance of 32.0 meters during the 4.0 s interval.