9)A car, moving along a straight stretch of highway, begins to accelerate at 0.0153 m/s
2
. It
takes the car 29.2 s to cover 1 km.
How fast was the car going when it first
began to accelerate?
Answer in units of m/s
s=v₀•t+a•t²/2
Solve for v₀
To find the initial speed of the car when it first began to accelerate, we can use the equation:
v = u + at
where:
v = final velocity
u = initial velocity (unknown)
a = acceleration
t = time
Given:
a = 0.0153 m/s^2
t = 29.2 s
Substituting the values into the equation:
0 = u + (0.0153 m/s^2) * (29.2 s)
Now, we can solve for u:
u = - (0.0153 m/s^2) * (29.2 s)
u = -0.44616 m/s^2
Therefore, the initial speed of the car when it first began to accelerate was approximately -0.44616 m/s.
To find the initial velocity of the car when it first began to accelerate, we can use the equation of motion:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
We are given that the acceleration is 0.0153 m/s^2 and the time taken to cover 1 km is 29.2 s.
First, we need to convert 1 km to meters:
1 km = 1000 meters
Substituting the given values into the equation, we have:
v = u + at
1000 = u + (0.0153)(29.2)
Now, we can solve for the initial velocity (u):
u = 1000 - (0.0153)(29.2)
Calculating this expression gives us the initial velocity of the car when it first began to accelerate.