A car accelerates uniformly at 3 m/s/s in order to overtake another car. If the initial speed was 30 m/s and the car accelerated until its speed was 40 m/s
–For how long did it accelerate? –How much distance was covered in that time?
v = Vi + 3 t
40 = 30 + 3 t
t = 10/3 seconds
d = 35 (10/3)
To find the time it took for the car to accelerate and the distance covered during that time, we can use the following kinematic equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
u = 30 m/s
a = 3 m/s^2
v = 40 m/s
1. Finding the time:
Substituting the given values into the equation, we have:
40 m/s = 30 m/s + 3 m/s^2 * t
Rearranging the equation to solve for t:
40 m/s - 30 m/s = 3 m/s^2 * t
10 m/s = 3 m/s^2 * t
Divide both sides of the equation by 3 m/s^2 to isolate t:
t = 10 m/s / 3 m/s^2
t ≈ 3.33 s
So, it took approximately 3.33 seconds for the car to accelerate.
2. Finding the distance:
To find the distance covered during this time, we can use another kinematic equation:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity
a = acceleration
t = time
Substituting the known values into the equation:
s = 30 m/s * 3.33 s + (1/2) * 3 m/s^2 * (3.33 s)^2
Evaluating the equation:
s ≈ 100 m + (1/2) * 3 m/s^2 * 11.09 s^2
s ≈ 100 m + 49.73 m
s ≈ 149.73 m
Therefore, the distance covered during the acceleration is approximately 149.73 meters.