A motor vehicle is uniformly retarded from a velocity 54km/h and finally come to a complete halt after 6s. Determine:
The rate at which it slowed down, and
The distance covered during this speed.
To determine the rate at which the motor vehicle slowed down, we need to calculate the deceleration.
The initial velocity (u) of the vehicle is 54 km/h, and the final velocity (v) is 0 km/h.
First, let's convert the velocities from km/h to m/s by using the conversion factor 1 km/h = 1000 m/3600 s:
Initial velocity (u) = 54 km/h = (54 * 1000) / 3600 m/s = 15 m/s
Final velocity (v) = 0 km/h = (0 * 1000) / 3600 m/s = 0 m/s
The time taken (t) for the vehicle to come to a complete halt is 6 s.
We can use the equation of motion to find the deceleration (a):
v = u + at
Substituting the values we know:
0 = 15 + a * 6
Rearranging the equation, we get:
6a = -15
Dividing both sides by 6, we find:
a = -15 / 6
Therefore, the rate at which the motor vehicle slowed down is approximately -2.5 m/s^2. The negative sign indicates that it is decelerating or slowing down.
To determine the distance covered during this deceleration, we can use another equation of motion:
s = ut + (1/2) * a * t^2
Substituting the values we know:
s = 15 * 6 + (1/2) * (-2.5) * 6^2
Simplifying the equation:
s = 90 - 45
Therefore, the distance covered during this deceleration is 45 meters.
To determine the rate at which the motor vehicle slowed down and the distance covered during this speed, we can use the equations of motion.
1. First, let's convert the initial velocity from km/h to m/s:
Initial velocity, u = 54 km/h = 54 * (1000/3600) m/s = 15 m/s
2. We need to find the acceleration (rate at which it slowed down) of the vehicle. The equation that links acceleration (a), initial velocity (u), final velocity (v), and time (t) is:
v = u + at
Since the final velocity is 0 m/s (the vehicle comes to a complete halt), the equation becomes:
0 = 15 + a * 6
Solving for a, we have:
a = -15/6 m/s²
(a negative value indicates retardation or deceleration)
Therefore, the rate at which the motor vehicle slowed down is -15/6 m/s² (or approximately -2.5 m/s²).
3. To find the distance covered during this speed, we can use another equation of motion:
s = ut + (1/2)at²
Substituting the values we know:
s = 15 * 6 + (1/2) * (-15/6) * (6^2)
s = 90 - 15
s = 75 meters
Therefore, the distance covered during this speed is 75 meters.
a = -54km/hr / 6s = -9 km/hr/s
s = vt + 1/2 at^2
= (54km/hr)(6/3600 hr) - 1/2 (9km/hr/s)(6s)(6/3600 hr)
= 54/600 km - 4.5(6)(1/600) km
= 9/200
= 0.045km = 450m