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Chapter 2: Problem 37

A car is speeding up and has an instantaneous velocity of $1.0 \mathrm{m} / \mathrm{s}\( in the \)+x\( -direction when a stopwatch reads \)10.0 \mathrm{s} .$ It has a constant acceleration of \(2.0 \mathrm{m} / \mathrm{s}^{2}\) in the \(+x\) -direction. (a) What change in speed occurs between \(t=10.0 \mathrm{s}\) and \(t=12.0 \mathrm{s} ?(\mathrm{b})\) What is the speed when the stopwatch reads \(12.0 \mathrm{s} ?\)

Short Answer

Expert verified
Answer: The change in speed between t1 and t2 is 4.0 m/s, and the speed of the car when the stopwatch reads t2 is 5.0 m/s.

Step by step solution

01

Calculate the acceleration time

Calculate the time interval during which the acceleration takes place as t鈧� - t鈧�: \(t = t鈧� - t鈧� = 12\,\text{s} - 10\,\text{s} = 2\,\text{s}\) The car is accelerating for 2 seconds.
02

Calculate the change in speed

To find the change in speed between t鈧� and t鈧�, we will use the first kinematic equation: \(v = u + at\) We will plug in the given values and solve the equation for the final velocity v: \(v = 1.0\,\text{m/s} + (2.0\,\text{m/s}^2)(2\,\text{s})\) Calculate the final velocity: \(v = 1.0\,\text{m/s} + (2.0\,\text{m/s}^2)(2\,\text{s}) = 1.0\,\text{m/s} + 4.0\,\text{m/s} = 5.0\,\text{m/s}\) Now, find the change in speed by subtracting the initial speed from the final speed: \(\Delta v = v - u = 5.0\,\text{m/s} - 1.0\,\text{m/s} = 4.0\,\text{m/s}\) The change in speed between t鈧� and t鈧� is 4.0 m/s.
03

Determine the speed at t=12 s

In part (b), we have to find the speed of the car when the stopwatch reads t鈧�. As we have already calculated the final speed (v) in Step 2, we know that the speed of the car at t=12 s is 5.0 m/s. So, the speed of the car when the stopwatch reads 12.0 s is 5.0 m/s.

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