5. A car moving at +12.0 m/s at the bottom of a hill, coasts up the hill with a uniform acceleration rate of -1.60 m/s'.
a. What is its displacement after 6.00 seconds?
b. What is its displacement after 9.00 seconds?
To solve these problems, we can use the equations of motion:
1. Displacement equation:
Δx = v₀t + 0.5at²
where Δx is displacement, v₀ is initial velocity, t is time, and a is acceleration.
Given:
Initial velocity, v₀ = +12.0 m/s
Acceleration, a = -1.60 m/s²
a. Displacement after 6.00 seconds:
t = 6.00 s
Δx = (12.0 m/s)(6.00 s) + 0.5(-1.60 m/s²)(6.00 s)²
= 72.0 m - 0.5(1.60 m/s²)(36.0 s²)
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= 72.0 m - 0.5
To solve this problem, we can use the kinematic equation:
\[ \text{displacement} = \text{initial velocity} \cdot \text{time} + \frac{1}{2} \cdot \text{acceleration} \cdot \text{time}^2 \]
Given:
Initial velocity (\( v_0 \)) = +12.0 m/s
Acceleration (\( a \)) = -1.60 m/s²
Time (\( t \)) = 6.00 s
a. To find the displacement after 6.00 seconds:
\[ \text{displacement} = (+12.0 \, \text{m/s}) \cdot (6.00 \, \text{s}) + \frac{1}{2} \cdot (-1.60 \, \text{m/s²}) \cdot (6.00 \, \text{s})^2 \]
Calculating:
\[ \text{displacement} = 72.0 \, \text{m} - 28.8 \, \text{m} \]
\[ \text{displacement} = 43.2 \, \text{m} \]
Therefore, the car's displacement after 6.00 seconds is 43.2 meters.
b. To find the displacement after 9.00 seconds:
\[ \text{displacement} = (+12.0 \, \text{m/s}) \cdot (9.00 \, \text{s}) + \frac{1}{2} \cdot (-1.60 \, \text{m/s²}) \cdot (9.00 \, \text{s})^2 \]
Calculating:
\[ \text{displacement} = 108.0 \, \text{m} - 64.8 \, \text{m} \]
\[ \text{displacement} = 43.2 \, \text{m} \]
Therefore, the car's displacement after 9.00 seconds is also 43.2 meters.