Starting from rest, a car takes 2.4 s to travel the first 15 m. Assuming a constant acceleration, how long will it take the car
1 answer:
Answer:
The time it will take the car to travel the next 15 m is 1 s.
Explanation:
Given;
initial velocity of the car, u = 0
time of motion, t = 2.4 s
initial distance traveled, d = 15 m
determine the constant acceleration of the car;
d = ut + ¹/₂at²
15 = 0 + (0.5 x 2.4²)a
15 = 2.88a
a = 15 / 2.88
a = 5.21 m/s²
The velocity of the car at the end of the first 15 m
v - u + at
v = 0 + 5.21 x 2.4
v = 12.5 m/s
The time to cover the next 15 m;
d₂ = vt + ¹/₂at²
15 = 12.5t + (0.5 x 5.21)t²
15 = 12.5t + 2.61t²
0 = 2.61t² + 12.5t - 15
Use the formula method to solve the above quadratic equation.
a = 2.61
b = 12.5
c = -15
Therefore, the time it will take the car to travel the next 15 m is 1 s.
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21) Acceleration from D to E:
22) The acceleration of the bus from D to E is
Explanation:
21)
The acceleration of an object is equal to the rate of change of velocity of the object. Mathematically:
where
u is the initial velocity
v is the final velocity
t is the time elapsed
In this problem, we want to measure the acceleration of the bus from point D to point E. We have:
- Initial velocity at point D: u = 0
- Final velocity at point E: v = 5 m/s
- Time elapsed from D to E: t = 21 - 16 = 5 s
Therefore, the acceleration between D and E is
22) This question is the same as 21), so the result is the same.
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4. The Coyote has an initial position vector of .
4a. The Coyote has an initial velocity vector of . His position at time
is given by the vector
where is the Coyote's acceleration vector at time
. He experiences acceleration only in the downward direction because of gravity, and in particular
where
. Splitting up the position vector into components, we have
with
The Coyote hits the ground when :
4b. Here we evaluate at the time found in (4a).
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