A car takes 11 s to go from v=0m/s to v = 22 m/s at constant acceleration.
If you wish to find the distance traveled using the equation d=1/2at2, what value should you use for a?
Please help on this i don't know what i should put i think m/s squared but i am not sure
Express your answer to two significant figures and include the appropriate units.
To find the value of acceleration (a) in the equation d=1/2at^2, we can rearrange the equation as:
d = (1/2) * a * t^2
Given that the car's initial velocity (v_i) is 0 m/s, the final velocity (v_f) is 22 m/s, and the time taken (t) is 11 s, we can use the equation v_f = v_i + at.
Substituting the values, we have:
22 m/s = 0 m/s + a * 11 s
Simplifying the equation, we get:
22 m/s = 11a
Dividing both sides by 11, we find:
a = 22 m/s / 11
Thus, the value of acceleration (a) is 2 m/s^2 (meters per second squared).
To find the distance traveled using the equation d = (1/2)at^2, you need to determine the value of acceleration (a) for the given scenario.
Since the car undergoes constant acceleration, we can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Given that the initial velocity (u) is 0 m/s, the final velocity (v) is 22 m/s, and the time (t) is 11 s, we can rearrange the formula as follows:
v = u + at
22 m/s = 0 m/s + a(11 s)
Simplifying the equation, we have:
22 m/s = 11a
Now, isolate the acceleration (a) by dividing both sides of the equation by 11:
a = 22 m/s / 11
a = 2 m/s^2
Therefore, the value of acceleration (a) for the given scenario is 2 m/s^2.
Now, you can substitute this value of acceleration into the formula d = (1/2)at^2 to find the distance traveled:
d = (1/2)(2 m/s^2)(11 s)^2
d = (1/2)(2 m/s^2)(121 s^2)
d = 1 m/s^2 * 121 s^2
d = 121 m
Thus, the distance traveled by the car is 121 meters.
10
a = (V-Vo)/t m/s^2.
V = 22 m/s.
Vo = 0.
t = 11 s.