A commuter backs her car out of her garage with an acceleration of 1.40m/s². (a) how long does it take her to reach a speed of 2.00m/s? (b) if she than brakes to a stop in 0.800s, what is her acceleration?
(a) To find the time taken to reach a speed of 2.00 m/s, we can use the equation:
v = u + at
Where:
v = final velocity = 2.00 m/s
u = initial velocity = 0 (since she starts from rest)
a = acceleration = 1.40 m/s²
t = time taken (unknown)
Rearranging the equation, we get:
t = (v - u) / a
t = (2.00 - 0) / 1.40
t = 2.00 / 1.40
t ≈ 1.43 seconds
So, it takes her approximately 1.43 seconds to reach a speed of 2.00 m/s.
(b) To find her acceleration when she brakes to a stop, we can use the equation:
v = u + at
Where:
v = final velocity = 0 (since she comes to a stop)
u = initial velocity = 2.00 m/s
a = acceleration (unknown)
t = time taken = 0.800 s
Rearranging the equation, we get:
a = (v - u) / t
a = (0 - 2.00) / 0.800
a = -2.00 / 0.800
a ≈ -2.50 m/s²
Her acceleration when braking to a stop is approximately -2.50 m/s² (negative because it is deceleration).
To solve these problems, we can use the equations of motion.
(a) To find the time it takes for the car to reach a speed of 2.00 m/s, we can use the equation:
v = u + at
Where:
v = final velocity (2.00 m/s)
u = initial velocity (0 m/s, as the car starts from rest)
a = acceleration (1.40 m/s²)
t = time
Rearranging the equation to solve for time (t), we have:
t = (v - u) / a
Substituting the given values:
t = (2.00 m/s - 0 m/s) / 1.40 m/s²
t = 2.00 m/s / 1.40 m/s²
t ≈ 1.43 s
Therefore, it takes her about 1.43 seconds to reach a speed of 2.00 m/s.
(b) To find the acceleration when she brakes to a stop in 0.800 s, we can use the equation:
v = u + at
Again, since the final velocity (v) is 0 m/s (as she comes to a stop), and the initial velocity (u) is 2.00 m/s, the equation becomes:
0 = 2.00 m/s + a * 0.800 s
Solving for acceleration (a), we have:
a = -u / t
a = -2.00 m/s / 0.800 s
a ≈ -2.50 m/s²
Therefore, her acceleration, when she brakes to a stop in 0.800 s, is approximately -2.50 m/s² (negative sign indicates deceleration).
To find the answers to these questions, we can use the equations of motion. The first equation relates displacement, initial velocity, final velocity, acceleration, and time:
Vf = Vi + at
Where:
Vf = final velocity (2.00 m/s in this case, part a)
Vi = initial velocity (0 m/s since she is starting from rest)
a = acceleration (1.40 m/s²)
t = time
(a) To find the time it takes for her to reach a speed of 2.00 m/s, we can rearrange the equation to solve for time:
t = (Vf - Vi) / a
Plugging in the values, we get:
t = (2.00 m/s - 0 m/s) / 1.40 m/s²
t = 2.00 s / 1.40 m/s²
t ≈ 1.43 s
So it takes her approximately 1.43 seconds to reach a speed of 2.00 m/s.
(b) To find her acceleration when she brakes to a stop, we can use the second equation of motion:
Vf = Vi + at
Where:
Vf = final velocity (0 m/s since she comes to a stop, part b)
Vi = initial velocity (2.00 m/s in this case, from part a)
a = acceleration (unknown, what we need to find)
t = time (0.800 s)
Rearranging the equation to solve for acceleration:
a = (Vf - Vi) / t
Plugging in the values we know:
a = (0 m/s - 2.00 m/s) / 0.800 s
a = -2.00 m/s / 0.800 s
a = -2.50 m/s²
So her acceleration while braking is -2.50 m/s², indicating that she is decelerating at a rate of 2.50 m/s². The negative sign indicates that the acceleration is in the opposite direction of the initial velocity.