Let’s solve each of the 4 kinematics problems step by step using the given equations:
---
Problem 1:
> A car begins at rest at a stoplight. After the light turns green, the car begins to accelerate 5 m/s² over the span of 3.5 seconds. What is the car’s displacement over this period of time?
Given:
- Initial velocity, \( v_i = 0 \) m/s (starts from rest)
- Acceleration, \( a = 5 \) m/s²
- Time, \( t = 3.5 \) s
- We need displacement, \( d \)
Use equation:
\[
d = v_i t + \frac{1}{2} a t^2
\]
Plug in values:
\[
d = (0)(3.5) + \frac{1}{2}(5)(3.5)^2
\]
\[
d = 0 + \frac{1}{2}(5)(12.25)
\]
\[
d = \frac{1}{2} \times 61.25 = 30.625 \text{ meters}
\]
✔ Answer: 30.625 meters
---
Problem 2:
> A cyclist is riding along at a speed of 12 m/s when she decides to come to a stop. The cyclist applies the brakes, decelerating her at a rate of 2.5 m/s² over the span of 5 seconds. What distance does she travel over this period of time?
Given:
- Initial velocity, \( v_i = 12 \) m/s
- Acceleration, \( a = -2.5 \) m/s² (negative because it’s deceleration)
- Time, \( t = 5 \) s
- Find displacement, \( d \)
Use equation:
\[
d = v_i t + \frac{1}{2} a t^2
\]
Plug in values:
\[
d = (12)(5) + \frac{1}{2}(-2.5)(5)^2
\]
\[
d = 60 + \frac{1}{2}(-2.5)(25)
\]
\[
d = 60 + (-31.25) = 28.75 \text{ meters}
\]
✔ Answer: 28.75 meters
---
Problem 3:
> A penny is dropped from the top of the Empire State Building. If the acceleration due to gravity is -9.8 m/s², and the Empire State Building is 381 meters tall, how long does it take for the penny to hit the ground?
Given:
- Initial velocity, \( v_i = 0 \) m/s (dropped, not thrown)
- Acceleration, \( a = -9.8 \) m/s² (downward, so negative if we define up as positive — but since it’s falling down, displacement will also be negative)
- Displacement, \( d = -381 \) m (since it falls downward from top to ground)
- Find time, \( t \)
Use equation:
\[
d = v_i t + \frac{1}{2} a t^2
\]
Plug in values:
\[
-381 = 0 \cdot t + \frac{1}{2}(-9.8)t^2
\]
\[
-381 = -4.9 t^2
\]
Divide both sides by -4.9:
\[
t^2 = \frac{381}{4.9} ≈ 77.755
\]
Take square root:
\[
t ≈ \sqrt{77.755} ≈ 8.82 \text{ seconds}
\]
✔ Answer: Approximately 8.82 seconds
---
Problem 4:
> A car traveling at 27 m/s slams on its brakes to come to a stop. It decelerates at a rate of 8 m/s². What is the stopping distance of the car?
Given:
- Initial velocity, \( v_i = 27 \) m/s
- Final velocity, \( v_f = 0 \) m/s (comes to stop)
- Acceleration, \( a = -8 \) m/s² (deceleration)
- Find displacement, \( d \)
Use equation:
\[
v_f^2 = v_i^2 + 2 a d
\]
Plug in values:
\[
0^2 = (27)^2 + 2(-8)d
\]
\[
0 = 729 - 16d
\]
Solve for \( d \):
\[
16d = 729
\]
\[
d = \frac{729}{16} = 45.5625 \text{ meters}
\]
✔ Answer: 45.5625 meters
---
##
✔ Final Answers Summary:
1.
30.625 meters
2.
28.75 meters
3.
≈8.82 seconds
4.
45.5625 meters
These solutions use the standard kinematic equations for constant acceleration. Always pay attention to signs (positive/negative) based on direction!
Parent Tip: Review the logic above to help your child master the concept of kinematics worksheet.
