A car traveling at 10 m/s accelerates at 2.03 m/s2 for 11 seconds. To the nearest meter how far does it travel?
To the nearest tenth of a m/s, what is the final velocity of the car in the previous problem?
Use the appropriate formulas.
X = (1/2) a t^2 + Vo*t
V = Vo + a*t
Vo is the initial velocity, 10 m/s
t = 11 s
a = 2.03 m/s^2
To find the distance travelled by the car, we can use the formula:
distance = initial velocity × time + 0.5 × acceleration × time^2
Given:
Initial velocity (u) = 10 m/s,
Acceleration (a) = 2.03 m/s^2,
Time (t) = 11 seconds.
Plugging the values into the formula:
distance = 10 m/s × 11 s + 0.5 × 2.03 m/s^2 × (11 s)^2
Calculating this expression will give us the distance travelled by the car. Let's do the math:
distance = 110 m + 0.5 × 2.03 m/s^2 × 121 s^2
distance = 110 m + 122.83 m
distance ≈ 232.83 m
Therefore, the car travels approximately 232.83 meters.
Now let's find the final velocity of the car. We can use the formula:
final velocity = initial velocity + acceleration × time
Given:
Initial velocity (u) = 10 m/s,
Acceleration (a) = 2.03 m/s^2,
Time (t) = 11 seconds.
Plugging the values into the formula:
final velocity = 10 m/s + 2.03 m/s^2 × 11 s
Calculating this expression will give us the final velocity of the car. Let's do the math:
final velocity = 10 m/s + (2.03 m/s^2 × 11 s)
final velocity = 10 m/s + 22.33 m/s
final velocity ≈ 32.33 m/s
Therefore, the final velocity of the car, rounded to the nearest tenth of a m/s, is approximately 32.3 m/s.