# 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?

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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?

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?

##### 1 Answer

#### Explanation:

Burt's acceleration tells you the rate at which its velocity is changing **with every passing second**.

You know that Burt started from rest, i.e. his initial velocity was *change* in its velocity divided by the *change* in time

#a = (Deltav)/(Deltat)#

Now, Burt's velocity is given to you in *miles per hour*, so make sure that you convert this to *miles per second* before figuring out the acceleration

#60.0 "mi"/color(red)(cancel(color(black)("h"))) * (1color(red)(cancel(color(black)("h"))))/"3600 s" = 1/60.0# #"mi s"^(-1)#

This means that Burt had an acceleration of

#a = (1/60.0 "mi s"^(-1) - "0.00 mi s"^(-1))/("4.00 s " - " 0.00 s") = "0.0041667 mi s"^(-2)#

This means that Burt's velocity **increases** by **with every passing second** he is in motion.

Now, you know that the final velocity reached during the event was

#183.58 "mi"/color(red)(cancel(color(black)("h"))) * (1color(red)(cancel(color(black)("h"))))/"3600 s" = "0.050944 mi s"^(-1)#

Your goal now is to figure out how much time passes until Burt reaches his top velocity. To do that, use the equation

#color(blue)(ul(color(black)(v_"final" = v_"initial" + a * t)))#

You know that Burt's acceleration *does not* change until he reaches his max velocity, so you can take the initial velocity to be

Rearrange to solve for

#t = (v_"final" - v_"initial")/a#

Plug in your values to find

#t = (0.050944 color(red)(cancel(color(black)("mi"))) color(red)(cancel(color(black)("s"^(-1)))) - 0.00 color(red)(cancel(color(black)("mi"))) color(red)(cancel(color(black)("s"^(-1)))))/(0.0041667color(red)(cancel(color(black)("mi")))"s"^color(red)(cancel(color(black)(-2))))#

#t = "12.226 s"#

Next, use this to figure out the **distance** covered while moving with uniform acceleration. To do that, sue the equation

#color(blue)(ul(color(black)(d = v_"initial" * t + 1/2 * a * t^2)))#

Once again, you can take

#d = 1/2 * "0.0041667 mi" * color(red)(cancel(color(black)("s"^(-2)))) * (12.226)^2 color(red)(cancel(color(black)("s"^2)))#

#d = "0.3114 mi"#

This means that the remaining distance

#d_"constant" = "5.00 mi" - "0.3114 mi" = "4.69 mi"#

was covered with a **constant velocity** of

#t = (4.69 color(red)(cancel(color(black)("mi"))))/(0.050944color(red)(cancel(color(black)("mi"))) "s"^(-1)) = "92.06 s"#

Therefore, you can say that Burt completed the course in

#t_"total" = "12.226 s" + "92.06 s" = color(darkgreen)(ul(color(black)("104 s")))#

The answer is rounded to three **sig figs**.