a) use v=u +at solve for a
b) use v^2=u^2+2as solve for s.
v is final vel; u is initial vel; t is time and a is accel. S is displacement; or distance in this case.
(a) Determine the acceleration of the car.
(b) How far does the car travel after acceleration starts?
b) use v^2=u^2+2as solve for s.
v is final vel; u is initial vel; t is time and a is accel. S is displacement; or distance in this case.
(b) Now, let's calculate how far our not-so-speedy racer travels after the acceleration starts. We can use the equation d = Vit + 0.5at^2 to find the distance. Plugging in the values, we have d = 40 * 5.0 + 0.5 * 8 * (5.0)^2. Doing the math, we get d = 200 + 100. Since clowns are bad at math, we'll just use a calculator. *beep boop beep* It turns out that the car travels a distance of 300 meters after the acceleration starts.
So, the acceleration of the car is 8 m/s^2 and it travels 300 meters after the acceleration starts. Keep in mind that these calculations assume constant acceleration and neglect other factors like friction, air resistance, and the presence of any creepy clowns on the race track!
v = u + at
where:
v is the final velocity (in this case, 0 m/s since the car comes to rest),
u is the initial velocity,
a is the acceleration, and
t is the time.
(a) To find the acceleration of the car, we can rearrange the above equation to solve for a:
a = (v - u) / t
Initial velocity, u = 40 m/s
Final velocity, v = 0 m/s
Time, t = 5.0 s
Plugging in the values, we get:
a = (0 - 40) / 5.0
a = -40 / 5.0
a = -8.0 m/s^2
Therefore, the acceleration of the car is -8.0 m/s^2.
(b) To find the distance traveled by the car after the acceleration starts, we can use the equation:
s = ut + (1/2)at^2
where:
s is the distance,
u is the initial velocity,
a is the acceleration, and
t is the time.
Plugging in the values, we get:
s = (40)(5.0) + (1/2)(-8.0)(5.0)^2
s = 200 + (1/2)(-8.0)(25)
s = 200 - 100
s = 100 m
Therefore, the car travels 100 meters after the acceleration starts.
(a) To determine the acceleration of the car, we can use the equation:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the final velocity, v, is 0 m/s (since the car comes to rest), the initial velocity, u, is 40 m/s, and the time, t, is 5.0 s. Plugging these values into the equation, we have:
0 = 40 + a * 5.0.
To solve for acceleration, we rearrange the equation:
a * 5.0 = -40.
Dividing both sides by 5.0:
a = -40 / 5.0 = -8.0 m/s^2.
Therefore, the acceleration of the car is -8.0 m/s^2.
(b) To find the distance traveled by the car after acceleration starts, we can use the equation:
d = ut + (1/2)at^2,
where d is the distance, u is the initial velocity, t is the time, and a is the acceleration.
Since we want to calculate the distance after the acceleration starts, we can use the time after the acceleration, which is 5.0 s.
Plugging in the values, we have:
d = (40 * 5.0) + (1/2)(-8.0)(5.0)^2.
Simplifying the equation, we get:
d = 200 - 100 = 100 m.
Therefore, the car travels a distance of 100 meters after the acceleration starts.