Additional Tutorial 3 2015
UNIVERSITI TUNKU ABDUL RAHMAN
Centre
:
Centre for Foundation Studies (CFS)
Unit Code
:
FHSP 1014
Course
:
Foundation in Science
Unit Title
:
Physics I
Year/ Trimester Session
:
:
Year 1 / Trimester 1
2015/05
Lecturer
:
Mr. Chua Lai Choy
Mr. Chin Kong Yew
Additional Questions 3: Kinematics
1. A balloon is 30.0 m above the ground and is rising vertically with a uniform speed when a coin is dropped from it. If the coin reaches the ground in 4.00 s, what is the speed of the balloon? Solution:- Initial velocity of coin = speed of balloon, v. by using the equation [Answer: 12.1 ms–1]
2. A car and train moves together along two parallel paths at 25.0 ms–1. The car then undergoes a uniform acceleration of -2.5 ms–2 because of a red light and comes to rest. It remains at rest for 45.0 s, then accelerates back to a speed of 25 m s – 1 at a rate of +2.5 ms–2. How far behind the train is the car when it reaches the speed of 25 ms–1, assuming that the train’s speed has remained constant at 25 ms–1. Solution:- For the car to stop we used the equation v2=v02 + 2as and v = v0 + at and For the car to speed up again, and time taken, Total distance moved by car in that time = 125 m + 125 m = 250 m. Total distance travelled by the train = 25 × (10+45+10) = 1625 m Therefore the car is (1625 – 250) = 1375 m behind the train. [Answer: 1375 m]
*3. A fugitive tries to hop on a freight train traveling at a constant speed of 6.0 m/s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 4.0 m/s2 to his maximum speed of 8.0 m/s. (a) How long does it take for him to catch up to the empty box car? (b) What is the distance traveled to reach the box car? Solution (a) acceleration of criminal Final velocity of criminal Velocity of train, Time taken for criminal to accelerate to 8 m/s from rest, Distance travelled by
Centre
:
Centre for Foundation Studies (CFS)
Unit Code
:
FHSP 1014
Course
:
Foundation in Science
Unit Title
:
Physics I
Year/ Trimester Session
:
:
Year 1 / Trimester 1
2015/05
Lecturer
:
Mr. Chua Lai Choy
Mr. Chin Kong Yew
Additional Questions 3: Kinematics
1. A balloon is 30.0 m above the ground and is rising vertically with a uniform speed when a coin is dropped from it. If the coin reaches the ground in 4.00 s, what is the speed of the balloon? Solution:- Initial velocity of coin = speed of balloon, v. by using the equation [Answer: 12.1 ms–1]
2. A car and train moves together along two parallel paths at 25.0 ms–1. The car then undergoes a uniform acceleration of -2.5 ms–2 because of a red light and comes to rest. It remains at rest for 45.0 s, then accelerates back to a speed of 25 m s – 1 at a rate of +2.5 ms–2. How far behind the train is the car when it reaches the speed of 25 ms–1, assuming that the train’s speed has remained constant at 25 ms–1. Solution:- For the car to stop we used the equation v2=v02 + 2as and v = v0 + at and For the car to speed up again, and time taken, Total distance moved by car in that time = 125 m + 125 m = 250 m. Total distance travelled by the train = 25 × (10+45+10) = 1625 m Therefore the car is (1625 – 250) = 1375 m behind the train. [Answer: 1375 m]
*3. A fugitive tries to hop on a freight train traveling at a constant speed of 6.0 m/s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 4.0 m/s2 to his maximum speed of 8.0 m/s. (a) How long does it take for him to catch up to the empty box car? (b) What is the distance traveled to reach the box car? Solution (a) acceleration of criminal Final velocity of criminal Velocity of train, Time taken for criminal to accelerate to 8 m/s from rest, Distance travelled by