a car moving at 30 m/s slows to a speed of 10 m/s in a time of 5 s. determine a.) the acceleration of the car (b). the distance it moves in the third second.
a. V = Vo + a*t
Vo = 30 m/s
V = 10 m/s
t = 5s.
Solve for a. It will be negative.
b. d = Vo*t + 0.5a*t^2
-4m/s
656
a. V = Vo + a*t
Vo = 30 m/s
V = 10 m/s
t = 5s.
Solve for a. It will be negative.
b. d = Vo*t + 0.5a*t^2
To determine the acceleration of the car, you can use the formula:
acceleration (a) = (final velocity - initial velocity) / time
In this case, the final velocity is 10 m/s, the initial velocity is 30 m/s, and the time is 5 seconds.
Using the given values, we can calculate the acceleration:
acceleration (a) = (10 m/s - 30 m/s) / 5 s
= (-20 m/s) / 5 s
= -4 m/s²
Therefore, the acceleration of the car is -4 m/s² (negative sign indicates deceleration).
To determine the distance the car moves in the third second, we need to calculate the distance traveled in the first three seconds and subtract the distance traveled in the first two seconds.
The distance traveled in the first three seconds can be calculated using the formula:
distance = initial velocity * time + (1/2) * acceleration * time²
Given:
Initial velocity = 30 m/s
Acceleration = -4 m/s²
Time = 3 seconds
distance = 30 m/s * 3 s + (1/2) * (-4 m/s²) * (3 s)²
= 90 m + (1/2) * (-4 m/s²) * 9 s²
= 90 m - 18 m
= 72 m
The distance traveled in the first two seconds can be calculated using the same formula with a time of 2 seconds:
distance = initial velocity * time + (1/2) * acceleration * time²
= 30 m/s * 2 s + (1/2) * (-4 m/s²) * (2 s)²
= 60 m + (1/2) * (-4 m/s²) * 4 s²
= 60 m - 8 m
= 52 m
Finally, to find the distance the car moves in the third second, we subtract the distance traveled in the first two seconds from the distance traveled in the first three seconds:
distance in the third second = 72 m - 52 m
= 20 m
Therefore, the car moves a distance of 20 meters in the third second.