A car accelerates from rest at -2.10 m/s2.
a. What is the velocity at the end of 4.0 s?
b. What is the displacement after 4.0 s?
v = a t = -2.1*4
x = .5 a t^2 = .5 (-2.1)(16)
Thank you!
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A particle accelerates from rest at 6.1 m/s
2
.
What is its speed 4.6 s after the particle
starts moving?
To find the answers to these questions, we need to use the equations of motion for uniformly accelerated motion. The key equation we will use is the second equation of motion, which relates the final velocity (v), initial velocity (u), acceleration (a), and time (t):
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Let's apply this equation to the given situation:
a. What is the velocity at the end of 4.0 s?
Given:
Initial velocity (u) = 0 m/s (since the car starts from rest)
Acceleration (a) = -2.10 m/s^2 (negative sign indicates the car is decelerating)
Time (t) = 4.0 s
Substituting these values into the equation, we have:
v = u + at
v = 0 + (-2.10) * 4.0
v = -8.40 m/s
Therefore, the velocity of the car at the end of 4.0 seconds is -8.40 m/s (negative sign indicates the car is moving in the opposite direction).
b. What is the displacement after 4.0 s?
To find the displacement, we can use the third equation of motion, which relates the displacement (s), initial velocity (u), acceleration (a), and time (t):
s = ut + (1/2)at^2
Given:
Initial velocity (u) = 0 m/s
Acceleration (a) = -2.10 m/s^2
Time (t) = 4.0 s
Substituting these values into the equation, we have:
s = ut + (1/2)at^2
s = 0 * 4.0 + (1/2) * (-2.10) * (4.0)^2
s = 0 + (1/2) * (-2.10) * 16.0
s = 0 + (-16.80)
s = -16.80 m
Therefore, the displacement of the car after 4.0 seconds is -16.80 meters (negative sign indicates the direction of motion).