A car travelling at 72km/h decelerates uniformly at 2m/s^2.Calculate(a)the distance it goes before it stops , (b)the time it takes to stop and (c)the distance it travels during the first and third seconds.
To calculate the distance the car goes before it stops, we can use the formula:
Distance = (Initial Velocity)^2 / (2 * Deceleration)
(a) Here, the initial velocity is given as 72 km/h. However, we need to convert it to m/s for consistency. 1 km/h is equal to (1/3.6) m/s. So, 72 km/h is equal to (72/3.6) m/s.
Initial Velocity = 72 km/h = (72/3.6) m/s = 20 m/s
Deceleration = 2 m/s^2
Now, we can plug these values into the formula:
Distance = (20 m/s)^2 / (2 * 2 m/s^2)
Distance = 200 m^2/s^2 / 4 m/s^2
Distance = 50 meters
Therefore, the distance the car goes before it stops is 50 meters.
(b) To calculate the time it takes to stop, we can use the formula:
Time = (Final Velocity - Initial Velocity) / Deceleration
In this case, the final velocity is 0 m/s (since the car stops), the initial velocity is 20 m/s, and the deceleration is 2 m/s^2:
Time = (0 m/s - 20 m/s) / 2 m/s^2
Time = -20 m/s / 2 m/s^2
Time = -10 seconds
Since time cannot be negative, we know there's an error in the calculation. However, let's assume that the time mentioned in the problem refers to the time taken to decelerate to 0 m/s.
In that case, the absolute value of the time would be 10 seconds.
(c) To calculate the distance traveled during the first and third seconds, we can make use of the following formulas:
Distance = Initial Velocity * Time + (1/2) * Deceleration * Time^2
For the first second (t = 1):
Distance1 = 20 m/s (Initial Velocity) * 1 second + (1/2) * 2 m/s^2 (Deceleration) * (1 second)^2
Distance1 = 20 m + (1/2) * 2 m/s^2
Distance1 = 20 m + 2 m
Distance1 = 22 meters
For the third second (t = 3):
Distance3 = 20 m/s (Initial Velocity) * 3 seconds + (1/2) * 2 m/s^2 (Deceleration) * (3 seconds)^2
Distance3 = 60 m + (1/2) * 2 m/s^2 * 9 s^2
Distance3 = 60 m + 9 m
Distance3 = 69 meters
Therefore, the distance the car travels during the first second is 22 meters, and during the third second is 69 meters.
Jak
c. d = Vo*t + 0.5a*t^2.
Vo = 20 m/s.
a = -2m/s^2
a. Vo = 72,000m/3600s. = 20 m/s.
V^2 = Vo^2 + 2a*d.
0 = 20^2 - 4*d.
b. V = Vo + a*t.
0 = 20 - 2t.
c.