A student driver backs his car out of his garage with an acceleration of 1.40m/s square. How long does it take him to reach a speed of 2.00m/s? and if he then breaks to a stop in 0.800s, what is his acceleration?
a. Vf=Vi+at
b. same
To find the time it takes for the student driver to reach a speed of 2.00 m/s, we can use the following equation of motion:
v = u + at
Where:
v = final velocity (2.00 m/s)
u = initial velocity (0 m/s)
a = acceleration (1.40 m/s²)
t = time
Now, let's rearrange the equation to solve for time (t):
t = (v - u) / a
Substituting the given values:
t = (2.00 m/s - 0 m/s) / 1.40 m/s²
Calculating this expression:
t = 2.00 m/s / 1.40 m/s²
t = 1.43 s
Therefore, it takes the student driver approximately 1.43 seconds to reach a speed of 2.00 m/s.
Next, to find the acceleration when the driver breaks to a stop, we can use the following equation of motion:
v = u + at
Where:
v = final velocity (0 m/s)
u = initial velocity (2.00 m/s)
a = acceleration (unknown)
t = time (0.800 s)
Rearranging the equation to solve for acceleration (a):
a = (v - u) / t
Substituting the given values:
a = (0 m/s - 2.00 m/s) / 0.800 s
Calculating this expression:
a = -2.00 m/s / 0.800 s
a = -2.50 m/s²
Therefore, the acceleration when the driver breaks to a stop is -2.50 m/s². The negative sign indicates deceleration or slowing down.