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Alchen [17]
3 years ago
11

A car traveling 75 km/h slows down at a constant 0.50 m/s2 just by "letting up on the gas." calculate (a) the distance the car c

oasts before it stops, (b) the time it takes to stop, and (c) the distance it travels during the first and fifth seconds.
Physics
1 answer:
Serjik [45]3 years ago
3 0

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.


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