A car of mass 1200kg travelling @ 72 km/h is brought to rest in 4s and has a deceleration of 5m/s find the distance moved during the deceleration
72 km/hr = 20.0 m/s
Average velocity during deceleration = 10 m/s.
10 m/s * 4 s = 40 meters.
The mass and deceleration rate data are not needed.
5m/s
To find the distance moved during deceleration, we can use the kinematic equation:
s = (v^2 - u^2) / (2a)
where:
s = distance moved during deceleration
v = final velocity (0 m/s since the car comes to rest)
u = initial velocity (72 km/h = 20 m/s)
a = deceleration (-5 m/s^2 since it's in the opposite direction)
First, let's convert the initial velocity from km/h to m/s:
u = 72 km/h * (1000 m/1 km) * (1 h/3600 s) = 20 m/s
Now, substitute the values into the equation and solve for s:
s = (0^2 - 20^2) / (2 * -5)
s = (0 - 400) / (-10)
s = 40
Therefore, the car moved a distance of 40 meters during the deceleration period.