A car starts from rest and travels for 5.3 s with a uniform acceleration of +1.2 m/s2. The driver then applies the brakes, causing a uniform acceleration of -2.3 m/s2. The breaks are applied for 1.60 s.
How fast is the car going at the end of the braking period?
How far has the car gone from its start?
VF=Vi + velocity gained-velocity lost
=0 + 1.2*5.3 + (-2.3)1.6
how far has it gone @bobpursley
James/Jahlen -- why did you switch names? Please keep the same name for all of your posts.
To find the final velocity of the car after the braking period, 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.
Given:
Initial velocity (u) = 0 m/s (since the car starts from rest)
Acceleration during braking (a) = -2.3 m/s^2
Time during braking (t) = 1.60 s
Using the formula, we can calculate the final velocity during braking:
v = u + at
v = 0 + (-2.3)(1.60)
v = -3.68 m/s
Therefore, the car is traveling at a speed of -3.68 m/s at the end of the braking period. The negative sign indicates that the car is moving in the opposite direction of its initial motion.
To find the distance traveled by the car from its start, we can use the equations of motion:
s = ut + (1/2)at^2
where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.
We need to calculate the total distance traveled during the acceleration phase and the braking phase separately.
Acceleration phase:
Initial velocity (u) = 0 m/s
Acceleration (a) = 1.2 m/s^2
Time (t) = 5.3 s
Using the formula, we can calculate the distance traveled during the acceleration phase:
s1 = ut + (1/2)at^2
s1 = 0 + (1/2)(1.2)(5.3)^2
s1 = 0 + (1/2)(1.2)(28.09)
s1 = 16.854 m
Braking phase:
Initial velocity (u) = -3.68 m/s (since the car is moving in the opposite direction)
Acceleration (a) = -2.3 m/s^2
Time (t) = 1.60 s
Using the formula, we can calculate the distance traveled during the braking phase:
s2 = ut + (1/2)at^2
s2 = (-3.68)(1.60) + (1/2)(-2.3)(1.60)^2
s2 = -5.888 + (1/2)(-2.3)(2.56)
s2 = -5.888 + (-2.592)
s2 = -8.480 m
Therefore, the car has traveled a total distance of s1 + s2 = 16.854 + (-8.480) = 8.374 m from its start. The negative sign indicates that the car has traveled in the opposite direction of its initial motion.