A car traveling at the speed of 30.0 m/sec encounters an emergency and comes to a complete stop. how much time will it take for the car to stop if its rate of deceleration is -4.0 m/sec
Using Calculus
a = -4
v = -4t + c
when t=0 , v = 30
30 = 0 + c
v = -4t + 30
s = -2t^2 + 30t + k
when t = 0 , s = 0, so k = 0
s = -2t^2 + 30t
when the car stops, v = 0
4t = 30
t = 30/4 = 7.5 seconds
when t = 7.5
s = -2(7.5)^2 + 30(7.5) = 112.5 m
To find the time it takes for the car to come to a complete stop, we can use the equation of motion:
v = u + at
Where:
v = final velocity (0 m/sec, since the car comes to a complete stop)
u = initial velocity (30.0 m/sec)
a = acceleration (deceleration in this case, which is -4.0 m/sec)
t = time
Rearranging the equation to solve for time (t), we get:
t = (v - u) / a
Substituting the given values into the equation:
t = (0 - 30.0) / -4.0
Simplifying the equation:
t = -30.0 / -4.0
t = 7.5 seconds
Therefore, it will take the car 7.5 seconds to come to a complete stop.
To find the time it takes for the car to stop, we can use the formula:
\(v = u + at\)
where:
- \(v\) is the final velocity (0 m/s, since the car comes to a complete stop),
- \(u\) is the initial velocity (30.0 m/s),
- \(a\) is the acceleration (deceleration in this case, -4.0 m/s²),
- \(t\) is the time.
Rearranging the formula to solve for \(t\), we have:
\(t = \frac{{v - u}}{a}\)
Substituting the given values into the formula, we get:
\(t = \frac{{0 - 30.0}}{-4.0}\)