Name: PHYSICS Unit 2c INMOTION Acceleration and Kinematic Equations gpb.org/physics-motion Practice Problems Date: Work each of the following problems. SHOW ALL WORK. 1. A sports car accelerates from rest to 26.8 m/s (roughly 60 mi/h) in 5.1 seconds. What is the acceleration of the car? 2. A child goes down a slide, starting from rest. If the length of the slide is 2 m and it takes the child 3 seconds to go down the slide, what is the child’s acceleration? 3. How far does a sled travel in 5 seconds while accelerating from 4 m/s to 10 m/s? 4. A fighter jet is catapulted off an aircraft carrier from rest to 75 m/s. If the aircraft carrier deck is 100 m long, what is the acceleration of the jet? PHYSICS Unit 2C Name: INMOTION>>>> Acceleration and Kinematic Equations gpb.org/physics-motion Practice Problems Date: Work each of the following problems. SHOW ALL WORK. 5. A driver notices an upcoming speed limit change from 45 mi/h (20m/s) to 25 mi/h (11 m/s). If she estimates the speed limit will change in 50 m, what acceleration is needed to reach the new speed limit before it begins? 6. One minute after takeoff, a rocket carrying the space shuttle into outer space reaches a speed of 447 m/s. What was the average acceleration of the rocket during that initial minute? 7. A sprinter accelerates from rest to a velocity of 12 m/s in the first 6 seconds of the 100-meter dash. a. How far does the sprinter travel during the first 6 seconds? b. How much farther does the sprinter have to travel to reach the finish line? PHYSICS Unit 2C Name: INMOTION>>> Acceleration and Kinematic Equations gpb.org/physics-motion Practice Problems Date: Work each of the following problems. SHOW ALL WORK. C. If the sprinter travels at a constant velocity of 12 m/s for the last 64 m, how long will it take to reach the finish line? 8. The school zone in front of your school has a posted speed limit of 25 mi/h, which is about 11 m/s. Let’s examine the stopping of a car in several different situations. a. The crossing guard holds up her stop sign, and the driver is paying attention well. The car moves at a constant velocity of 11 m/s for 2.3 seconds while the driver reacts, then slows down at a constant rate of -4.5 m/s? What is the stopping distance for the car in this situation? Name: Unit 2C PHYSICS INMOTION>> Acceleration and Kinematic Equations gpb.org/physics-motion Practice Problems Date: Work each of the following problems. SHOW ALL WORK. b. A child appears to be running into the street ahead. It takes 2.3 seconds for the driver to react and begin to brake, but this time at a rate of -7.5 m/s2. What is the stopping distance for the car in this situation? c. The driver is looking at her phone and has a total reaction time of 4.6 seconds as the car is moving at a constant speed of 11 m/s. If the driver slams on her brakes and slows down at a rate of -8.2 m/s?, what is the stopping distance for the car in this situation? >>> Acceleration and Kinematic Equations gpb.org/physics-motion Practice Problems Date: Work e”> >> Acceleration and Kinematic Equations gpb.org/physics-motion Practice Problems Date: Work ea”> > Acceleration and Kinematic Equations gpb.org/physics-motion Practice Problems Date: Work eac”>
The Correct Answer and Explanation is :
Here are the solutions and explanations for the given kinematics problems:
1. Car acceleration:
- Given: Initial velocity ( u = 0 ), final velocity ( v = 26.8 \, \text{m/s} ), time ( t = 5.1 \, \text{s} )
- Formula: ( a = \frac{v – u}{t} )
- Calculation: ( a = \frac{26.8 – 0}{5.1} = 5.25 \, \text{m/s}^2 )
- Answer: ( 5.25 \, \text{m/s}^2 )
2. Child on a slide:
- Given: ( u = 0 ), ( s = 2 \, \text{m} ), ( t = 3 \, \text{s} )
- Formula: ( s = ut + \frac{1}{2}at^2 )
- Calculation: ( 2 = 0 + \frac{1}{2}a(3)^2 )
( a = \frac{4}{9} = 0.444 \, \text{m/s}^2 ) - Answer: ( 0.444 \, \text{m/s}^2 )
3. Sled distance:
- Given: ( u = 4 \, \text{m/s} ), ( v = 10 \, \text{m/s} ), ( t = 5 \, \text{s} )
- Formula: ( s = ut + \frac{1}{2}at^2 ), where ( a = \frac{v – u}{t} = \frac{10 – 4}{5} = 1.2 )
- Calculation: ( s = 4(5) + \frac{1}{2}(1.2)(5)^2 = 20 + 15 = 35 \, \text{m} )
- Answer: ( 35 \, \text{m} )
4. Fighter jet acceleration:
- Given: ( u = 0 ), ( v = 75 \, \text{m/s} ), ( s = 100 \, \text{m} )
- Formula: ( v^2 = u^2 + 2as )
- Calculation: ( 75^2 = 0 + 2a(100) )
( a = \frac{5625}{200} = 28.125 \, \text{m/s}^2 ) - Answer: ( 28.125 \, \text{m/s}^2 )
5. Driver deceleration:
- Given: ( u = 20 \, \text{m/s} ), ( v = 11 \, \text{m/s} ), ( s = 50 \, \text{m} )
- Formula: ( v^2 = u^2 + 2as )
- Calculation: ( 11^2 = 20^2 + 2a(50) )
( a = \frac{121 – 400}{100} = -2.79 \, \text{m/s}^2 ) - Answer: ( -2.79 \, \text{m/s}^2 )
Explanation (Summary):
These problems use kinematic equations that describe motion under constant acceleration. The equations ( v = u + at ), ( s = ut + \frac{1}{2}at^2 ), and ( v^2 = u^2 + 2as ) are critical tools. For each problem:
- Identify the known values (initial velocity, time, etc.).
- Choose the appropriate equation based on the missing variable.
- Solve algebraically for the unknown.
Showing all work step-by-step ensures accurate results and highlights relationships between acceleration, velocity, and displacement. These equations are foundational for analyzing real-world scenarios involving motion.