A car starts from rest accelerate uniformly until it reaches a velocity of 30m/s after 5s.it traves with uniform velocity for 15s and is being brought to rest in 10 with a uniform retardation.dEtermine the following (a) the acceleration of the car.(b) the retardation. (c) the distance covered after 5s.(d) the total distance covered.
Velocity =30/5=6m/s, retardation =30/10=3m/s, distance=30x5=150m, distance = 30x30 =900m
a) & b)
v = at
solve for a
c) x = 1/2 at^2
d) middle:
x = vt
slowing
x = vt- 1/2 at^2
To solve this problem, we can use the equations of constant acceleration motion.
(a) To find the acceleration of the car, we can use the equation:
v = u + at
where:
v = final velocity = 30 m/s
u = initial velocity = 0 m/s (since the car starts from rest)
t = time = 5 s
Rearranging the equation, we have:
a = (v - u) / t
a = (30 - 0) / 5
a = 6 m/s^2
So, the acceleration of the car is 6 m/s^2.
(b) To determine the retardation, we can use another equation:
v = u + at
where:
v = final velocity = 0 m/s (since the car comes to rest)
u = initial velocity = 30 m/s (since the car was traveling with uniform velocity before decelerating)
t = time = 10 s
Rearranging the equation, we have:
a = (v - u) / t
a = (0 - 30) / 10
a = -3 m/s^2
Since the car is decelerating, the acceleration (retardation) is negative. Therefore, the retardation of the car is -3 m/s^2.
(c) To find the distance covered after 5 seconds, we can use the equation of motion:
s = ut + (1/2)at^2
where:
s = distance covered
u = initial velocity = 0 m/s (since the car starts from rest)
t = time = 5 s
a = acceleration = 6 m/s^2
Plugging in the values, we get:
s = (0 * 5) + (1/2) * 6 * (5^2)
s = 0 + (1/2) * 6 * 25
s = 0 + 75
s = 75 m
Therefore, the distance covered after 5 seconds is 75 meters.
(d) To find the total distance covered, we need to consider the three segments: acceleration, constant velocity, and deceleration.
The distance covered during acceleration can be found using the kinematic equation:
s = ut + (1/2)at^2
where:
u = initial velocity = 0 m/s
t = time for acceleration = 5 s
a = acceleration = 6 m/s^2
Plugging in the values, we find:
s1 = (0 * 5) + (1/2) * 6 * (5^2)
s1 = 0 + (1/2) * 6 * 25
s1 = 0 + 75
s1 = 75 m
The distance covered during constant velocity is calculated using:
s2 = velocity * time
s2 = 30 * 15
s2 = 450 m
The distance covered during deceleration can also be found using the kinematic equation:
s = ut + (1/2)at^2
where:
u = initial velocity = 30 m/s
t = time for deceleration = 10 s
a = retardation = -3 m/s^2
Plugging in the values, we find:
s3 = (30 * 10) + (1/2) * (-3) * (10^2)
s3 = 300 + (-1.5) * 100
s3 = 300 - 150
s3 = 150 m
To calculate the total distance covered, we sum up the distances covered during each segment:
Total distance = s1 + s2 + s3
Total distance = 75 + 450 + 150
Total distance = 675 meters
Therefore, the total distance covered by the car is 675 meters.