Solution:
At 1st convert km/h to m/s. 1 km = 1000 m, 1 h = 3600 s, 1 km/m = 1000/3600 = 5/18 m/s
Initial velocity = 75 * 5/18 = 20.8 m/s
The car’s velocity decreases from 20.8 m/s to 0 m/s at the rate of 0.5 m/s each second. We have the final velocity, initial velocity, and the acceleration.
Now according to the equation determine the distance.
vf^2 = vi^2 + 2 * a * d
a = -0.5 m/s^2
0 = 20.8^2 + 2 * -0.5 * d
so d = 431.64 m
since we have the final velocity, initial velocity, and the acceleration. Use the following equation to determine time.
vf = vi + a * t
0 = 20.8 – 0.5 * t
Solve for t = 41 seconds
(c) the distance travels by it during the first and fifth second are.
d = vi * t + ½ * a * t^2
d1 = 20.8 * 1 – ½ * 0.5 * 1^2 = 20.55 m
The easiest way to the distance for the 5th second is:
d = vi * t + ½ * a * t^2, a = -0.5
d5 = 20.8 * 5 – ½ * 0.5 * 5^2 = 91.5 m
d6 = 20.8 * 6 – ½ * 0.5 * 6^2 = 106.8m
d6 – d5 = 15.3 m
this is the required solution.
The distance covered in the fifth second is 23.05 m.
We have to convert the initial velocity to m/s as follows;
75 × 1000/3600 = 20.8 m/s
Given that;
v^2 = u^2 - 2as
When v = 0 m/s
u^2 = 2as
s = u^2/2a
s = (20.8 m/s)^2/2 × 0.50 m/s2
s = 432.6 m
The time taken to stop is obtained from;
v = u - at
v = 0
u = at
t = u/a = 20.8 m/s/ 0.50 m/s2
t = 41.6 seconds
Distance covered in the first second;
s = ut + 1/2at^2
s = 20.8(1) + 1/2 (0.50) (1)^2
s = 21.05 m
Distance covered after five seconds;
s = 20.8(5) + 1/2 (0.50)(5)^2
s = 104 + 6.25
s= 110.25 m
Distance covered after four seconds;
s = 20.8(4) + 1/2 (0.50)(4)^2
s = 83.2 + 4 =87.2 m
Distance covered in the fifth second = 110.25 m - 87.2 m = 23.05 m
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