<< Chapter < Page Chapter >> Page >

A manned rocket accelerates at a rate of 20 m/s 2 size 12{"20 m/s" rSup { size 8{2} } } {} during launch. How long does it take the rocket to reach a velocity of 400 m/s?

To answer this, choose an equation that allows you to solve for time t size 12{t} {} , given only a size 12{a} {} , v 0 size 12{v rSub { size 8{0} } } {} , and v size 12{v} {} .

v = v 0 + at size 12{v=v"" lSub { size 8{0} } + ital "at"} {}

Rearrange to solve for t size 12{t} {} .

t = v v 0 a = 400 m/s 0 m/s 20 m/s 2 = 20 s size 12{t= { {v - v"" lSub { size 8{0} } } over {a} } = { {"400 m/s" - "0 m/s"} over {"20 m/s" rSup { size 8{2} } } } ="20 s"} {}

Section summary

  • To simplify calculations we take acceleration to be constant, so that a - = a size 12{ { bar {a}}=a} {} at all times.
  • We also take initial time to be zero.
  • Initial position and velocity are given a subscript 0; final values have no subscript. Thus,
    Δ t = t Δ x = x x 0 Δ v = v v 0
  • The following kinematic equations for motion with constant a size 12{a} {} are useful:
    x = x 0 + v - t size 12{x=x rSub { size 8{0} } + { bar {v}}t} {}
    v - = v 0 + v 2 size 12{ { bar {v}}= { {v rSub { size 8{0} } +v} over {2} } } {}
    v = v 0 + at size 12{v=v rSub { size 8{0} } + ital "at"} {}
    x = x 0 + v 0 t + 1 2 at 2 size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {}
    v 2 = v 0 2 + 2 a x x 0 size 12{v rSup { size 8{2} } =v rSub { size 8{0} } rSup { size 8{2} } +2a left (x - x rSub { size 8{0} } right )} {}
  • In vertical motion, y size 12{y} {} is substituted for x size 12{x} {} .

Problems&Exercises

An Olympic-class sprinter starts a race with an acceleration of 4 . 50 m/s 2 size 12{4 "." "50 m/s" rSup { size 8{2} } } {} . (a) What is her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period.

(a) 10 . 8 m/s size 12{"10" "." 8" m/s"} {}

(b)

Line graph of position in meters versus time in seconds. The line begins at the origin and is concave up, with its slope increasing over time.

(a) A light-rail commuter train accelerates at a rate of 1 . 35 m/s 2 size 12{1 "." "35 m/s" rSup { size 8{2} } } {} . How long does it take to reach its top speed of 80.0 km/h, starting from rest? (b) The same train ordinarily decelerates at a rate of 1 . 65 m/s 2 size 12{1 "." "65 m/s" rSup { size 8{2} } } {} . How long does it take to come to a stop from its top speed? (c) In emergencies the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in m/s 2 size 12{"m/s" rSup { size 8{2} } } {} ?

(a) 16 . 5 s size 12{`"16" "." "5 s"} {}

(b) 13 . 5 s size 12{"13" "." "5 s"} {}

(c) 2 . 68 m/s 2 size 12{` - 2 "." "68 m/s" rSup { size 8{2} } } {}

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 3 . 33 × 10 2 s size 12{3 "." "33"´"10" rSup { size 8{-2} } " s"} {} , calculate the distance over which the puck accelerates.

0 . 799 m size 12{0 "." "799 m"} {}

A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of 0.250 m. (a) How long did the acceleration last? (b) Calculate the acceleration.

Professional Application:

A woodpecker’s brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker’s head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm. (a) Find the acceleration in m/s 2 and in multiples of g g = 9 . 80 m/s 2 size 12{g left (g=9 "." "80"" m/s" rSup { size 8{2} } right )} {} . (b) Calculate the stopping time. (c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain’s deceleration, expressed in multiples of g ?

An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration? (b) How long does the collision last?

(a) 80 . 4 m/s 2 size 12{ - "80" "." 4" m/s" rSup { size 8{2} } } {}

(b) 9 . 33 × 10 2 s size 12{9 "." "33" times "10" rSup { size 8{ - 2} } " s"} {}

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0 . 500 m/s 2 size 12{0 "." "500 m/s" rSup { size 8{2} } } {} for 7.00 s. (a) What is his final velocity? (b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save? (c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, with a maximum speed of 183.58 mi/h. The one-way course was 5.00 mi long. Acceleration rates are often described by the time it takes to reach 60.0 mi/h from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?

104 s

(a) A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration. (b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?

(a) v = 12 . 2 m/s size 12{v="12" "." "2 m/s"} {} ; a = 4 . 07 m/s 2 size 12{a=4 "." "06 m/s" rSup { size 8{2} } } {}

(b) v = 11 . 2 m/s size 12{v="11" "." "2 m/s"} {}

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Kinematics. OpenStax CNX. Sep 11, 2015 Download for free at https://legacy.cnx.org/content/col11878/1.5
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Kinematics' conversation and receive update notifications?

Ask