Freight trains can produce only relatively small accelerations and decelerations. (a) What is the final velocity of a freight train that accelerates at a rate of $latex 0.0500\:m/s^2&s=1&fg=000000$ for 8.00 minutes, starting with an initial velocity of 4.00 m/s? (b) If the train can slow down at a rate of $latex 0.550\:m/s&s=1&fg=000000$, how long will it take to … Continue reading College Physics Problem 2.29

# Tag: Speed

A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90 s. (a) What is its average acceleration? (b) How far does it travel in that time? SOLUTION: Part (a). The average acceleration of the motorcycle can be solved using the equation $latex \overline{a}=\frac{\Delta v}{\Delta t}&s=0&fg=000000$. Substitute the given into the equation. … Continue reading College Physics Problem 2.28

In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes $latex 3.33\times 10^{-2}\:s&s=1&fg=000000$, calculate the distance over which the puck accelerates. SOLUTION: The best equation that can be used to solve this problem is $latex \Delta{x}=v_{ave}t&s=0&fg=000000$. That is, $latex … Continue reading College Physics Problem 2.27

Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart. (a) Make a sketch of the solution. (b) List the knowns in this problem. (c) How long does the acceleration take? To solve this part, first identify the unknown, and then discuss how you chose the … Continue reading College Physics Problem 2.26

At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of $latex 2.00\:m/s^2&s=1&fg=000000$. (a) How far does she travel in the next 5.00 s? (b) What is her final velocity? (c) Evaluate the result. Does it make sense? Solution: Part (a). For this part, we use the … Continue reading College Physics Problem 2.25

An Olympic-class sprinter starts a race with an acceleration of $latex 4.50\:m/s^2&s=1&fg=000000$. (a) What is her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period. SOLUTION: We are given $latex a=4.50\:m/s^2&s=0&fg=000000$ $latex t=2.40\:s&s=0&fg=000000$ $latex v_0=0\:m/s&s=0&fg=000000$ a) The unknown is $latex v_f&s=1&fg=000000$. The formula in solving for $latex v_f&s=1&fg=000000$ is $latex … Continue reading College Physics Problem 2.20

Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed and time are classified). What is its average acceleration in $latex m/s^{2}&s=1&fg=000000$ and in multiples of $latex g\:(9.80\:m/s^{2})&s=1&fg=000000$ ? SOLUTION: The formula for acceleration is $latex a=\frac{v_f-v_0}{t}&s=1&fg=000000$ Substituting the given values $latex a=\frac{6.5\times 10^3\:m/s-0\:m/s}{60.0\:s}&s=1&fg=000000$ … Continue reading College Physics Problem 2.19

A commuter backs her car out of her garage with an acceleration of $latex 1.40\:m/s^{2}&s=1&fg=000000$ . (a) How long does it take her to reach a speed of 2.00 m/s? (b) If she then brakes to a stop in 0.800 s, what is her deceleration? SOLUTION: a) The formula for acceleration is $latex a=\frac{v_f-v_0}{t}&s=1&fg=000000$ If we … Continue reading College Physics Problem 2.18

Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! … Continue reading College Physics Problem 2.17

A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its acceleration? SOLUTION: The formula for acceleration is $latex a=\frac{change\:in\:velocity}{time}=\frac{\Delta v}{time}=\frac{v_f-v_0}{t}&s=1&fg=000000$ Substituting the given values $latex a=\frac{30.0\:\frac{m}{s}-0}{7.00\:s}&s=1&fg=000000$ $latex a=4.29\:m/s^{2}&s=1&fg=000000$ You can now buy the complete solution manual for College Physics. Just complete the google form below.