A car going at 30 m/s undergoes an acceleration of 2m/s squared for 4 seconds. What is it's final speed? How far did it travel while it was accelerating?
2 m/s^2=?-30m/s^2/4s
?-30=4*2
?-30=8
?=8+30=38
1st speed=38m/s
A certain airplane has an acceleration of 15.0 m/s2
How fast will it be moving2.5s after it starts down runway
How far down runway during the 2.5s
The first speed is 38m/s
To find the final speed of the car, you need to use the formula:
final speed = initial speed + (acceleration × time).
In this case, the initial speed is 30 m/s, the acceleration is 2 m/s^2, and the time is 4 seconds. Plugging in these values into the formula, we get:
final speed = 30 m/s + (2 m/s^2 × 4 s)
final speed = 30 m/s + 8 m/s
final speed = 38 m/s
Therefore, the final speed of the car is 38 m/s.
To find the distance traveled by the car while it was accelerating, you can use the formula:
distance = initial speed × time + (0.5 × acceleration × time^2).
Since the car starts from rest, the initial speed is 0 m/s. The acceleration is 2 m/s^2, and the time is 4 seconds. Plugging in these values into the formula, we get:
distance = 0 m/s × 4 s + (0.5 × 2 m/s^2 × (4 s)^2)
distance = 0 m + (0.5 × 2 m/s^2 × 16 s^2)
distance = 0 m + (0.5 × 2 × 16) m
distance = 0 m + 16 m
distance = 16 m
Therefore, the car traveled a distance of 16 meters while it was accelerating.
V(t) = Vo + a*t
X(t) = Vo*t + (1/2)a*t^2
Learn them and use them.
t = 4. Vo = 30 m/s a = 2 m/s^2
BTW, drop the apostrophe in it's.