1.
A car travels down a straight, flat, level road. Initially, the
car
is traveling at 80.0 km/hr
Northward
. After traveling 0.50
0
km
, the car is traveling at
95
.0 km/hr. What is the magn
i
tude
of the car
’
s acceleration
during this interval
?
To find the magnitude of the car's acceleration during this interval, we need to calculate the change in velocity and divide it by the time taken.
1. First, convert the initial and final velocities from km/hr to m/s. We know that 1 km/hr = 0.2778 m/s.
Initial velocity (v1) = 80.0 km/hr * 0.2778 m/s = 22.22 m/s
Final velocity (v2) = 95.0 km/hr * 0.2778 m/s = 26.39 m/s
2. Next, calculate the change in velocity (delta v).
delta v = v2 - v1 = 26.39 m/s - 22.22 m/s = 4.17 m/s
3. Now, calculate the time taken. We are given that the car has traveled a distance of 0.50 km.
Convert the distance from km to m. We know that 1 km = 1000 m.
Distance (d) = 0.50 km * 1000 m/km = 500 m
Given that the car has traveled this distance, we need to find the time taken (t) to travel this distance.
Use the formula:
distance = velocity * time
500 m = (v1 + v2) / 2 * t
Substituting the values, we get:
500 m = (22.22 m/s + 26.39 m/s) / 2 * t
Solve for t:
500 m = 24.31 m/s * t
t = 20.55 seconds
4. Finally, calculate the magnitude of the car's acceleration.
Acceleration (a) = delta v / t
a = 4.17 m/s / 20.55 s = 0.203 m/s^2
Therefore, the magnitude of the car's acceleration during this interval is 0.203 m/s^2.