Concept explainers
A car traveling in a straight line with an initial velocity of 10 m/s accelerates at a rate of 3.0 m/s2 to a velocity of 34 m/s.
- a. How much time does it take for the car to reach the velocity of 34 m/s?
- b. What is the distance covered by the car in this process?
- c. Compute values of the distance traveled at 1-second intervals and carefully draw a graph of distance plotted against time for this motion.
(a)
The time taken for the car to reach the velocity of 34 m/s.
Answer to Problem 4SP
The time taken for the car to reach the velocity of 34 m/s is
Explanation of Solution
Given info: Initial velocity of the car is 10 m/s, acceleration is
Write the expression for Newton’s equation of motion.
Here,
v is the final velocity
t is the initial time
a is the acceleration
Re-arrange the above equation to get t.
Substitute 34 m/s for v, 10 m/s for
Conclusion:
The time taken for the car to reach the velocity of 34 m/s is
(b)
The distance covered by the the car.
Answer to Problem 4SP
The distance covered by the car is
Explanation of Solution
Given info: Initial velocity of the car is 10 m/s, acceleration is
Write the expression for distance.
Here,
v is the final velocity
t is the initial time
s is the distance
Substitute 8s for t, 10 m/s for
The distance covered by the car is
(c)
To plot: The graph of distance versus time.
Answer to Problem 4SP
Distance is plotted against time in the graph.
Explanation of Solution
Write the expression for distance.
At 0s,
Substitute 0s for t, 10 m/s for
At 1s,
Substitute 1s for t, 10 m/s for
At 2s,
Substitute 2s for t, 10 m/s for
At 3s,
Substitute 3s for t, 10 m/s for
At 4s,
Substitute 4s for t, 10 m/s for
At 5s,
Substitute 5s for t, 10 m/s for
At 6s,
Substitute 6s for t, 10 m/s for
At 7s,
Substitute 7s for t, 10 m/s for
At 8s,
Substitute 8s for t, 10 m/s for
The above values are plotted in the below graph.
Conclusion:
Distance is plotted against time in the graph.
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Chapter 2 Solutions
Physics of Everyday Phenomena
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