A car accelerates at 2.25m/s^2 along a straight road. It passes two marks that are 30.5m apart at times t=3.80s and t=4.90s. What was the car's velocity at t=0?
Average speed between the marks = 30.5/1.10 = 27.73 m/s
Speed increase between marks = 2.25*1.1 = 2.475 m/s
Speed passing first mark 27.73 - 1.24 = 26.49 m/s
26.49 = Vo + 2.25*3.80
Solve for Vo
To find the car's velocity at t=0, we need to use the kinematic equation for acceleration:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, we are given the acceleration (a = 2.25 m/s^2) and the time at which two marks are passed (t=3.80s and t=4.90s) along with the distance between the marks (30.5m). Since we are looking for the velocity at t=0, we need to find the initial velocity (u).
We can start by finding the final velocity at t=4.90s using the formula:
v = u + at
Plugging in the values:
v = u + (2.25 m/s^2)(4.90s)
Now, we need to find the distance covered by the car between t=3.80s and t=4.90s. The distance (s) can be calculated using the equation:
s = ut + (1/2)at^2
Since the car is moving in a straight line, the distance covered is equal to the difference between the positions at t=4.90s and t=3.80s.
s = s2 - s1
Substituting the values:
s = (u)(4.90s) + (1/2)(2.25 m/s^2)(4.90s)^2 - (u)(3.80s) - (1/2)(2.25 m/s^2)(3.80s)^2
We know that s = 30.5m. Substituting this value:
30.5m = (u)(4.90s) + (1/2)(2.25 m/s^2)(4.90s)^2 - (u)(3.80s) - (1/2)(2.25 m/s^2)(3.80s)^2
Simplifying this equation will give us a value for u, which is the initial velocity at t=0. Once we solve for u, we will have the answer to the question.