# Video: Using Kinematic Equations to Find the Time Taken to Travel a Given Distance for Periods of Constant Acceleration

In 1967, New Zealander Burt Munro set the world record for the Indian motorcycle, on the Bonneville Salt Flats of Utah, of 295.38 km/h. The one-way course was 8.00 km long. Acceleration rates are often described by the time it takes to reach 96.0 km/h from rest. If this time was 4.00 s and Burt accelerated that this rate until he reached his maximum speed, how long did it take Burt to complete the course?

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### Video Transcript

In 1967, New Zealander Burt Munro set the world record for the Indian motorcycle on the Bonneville Salt Flats of Utah of 295.38 kilometers per hour. The one-way course was 8.00 kilometers long. Acceleration rates are often described by the time it takes to reach 96.0 kilometers per hour from rest. If this time was 4.00 seconds and Burt accelerated that this rate until he reached his maximum speed, how long did it take Burt to complete the course?

Let’s begin by highlighting some of the vital information we’re given. We’re told that Burt’s maximum speed was 295.38 kilometers per hour. The course he travelled was 8.00 kilometers long. We’re told that it took Burt 4.00 seconds in order to reach a speed of 96.0 kilometers per hour from rest. We want to solve for the total time it took Burt to travel a course. We’ll call that 𝑡.

Let’s begin by solving for Burt’s acceleration over the first four seconds of travel. Recall that acceleration is defined as the change in velocity divided by the change in time. In our case, we’re told that Burt was able to achieve a speed of 96.0 kilometers per hour in a time of 4.00 seconds. Before we calculate this acceleration, we want to convert the units of kilometers per hour to meters per second.

To do that, let’s recall that one kilometer is equal to 1000 meters and one hour is equal to 3600 seconds. So if we take our given speed of 96.0 kilometers per hour and multiply it by 1000 meters per 3600 seconds, then this multiplication will give us a result in units of meters per second. This is equal to 26.67 meters per second. Now we can insert this value for speed in meters per second into the numerator of our acceleration equation.

And when we calculate this fraction, we find a value of six and two-thirds meters per second squared, which we can abbreviate as 6.667. This is the acceleration that Burt experienced up until reaching the maximum speed, 𝑣 sub max. Since we want to solve for the time it took Burt to complete the whole 8.00 kilometer course, let’s find out how long it took him to reach a speed of 𝑣 max accelerating from rest at this rate.

Since we have a constant acceleration value, let’s refer to the kinematic equations to see if they can be useful to us. Remember that the kinematic equations are true whenever acceleration is constant, as it is in our case. Looking at the very first kinematic equation listed, we see it can be useful to us because 𝑣 sub 𝑓 in our case is 𝑣 sub max; 𝑣 sub zero is zero because Burt started from rest; 𝑎, we just solved for; and 𝑡 is what we want to find.

Since we don’t know whether the 𝑡 that were solving for is the total time it took Burt to complete the course, let’s provisionally call it 𝑡 sub one. If we divide both sides of this equation by 𝑎, then the acceleration term on the right cancels out and we find that 𝑡 sub one is equal to the 𝑣 max divided by 𝑎. We know 𝑎; we solved for that as 6.667 meters per second squared. And 𝑣 max is given to us as 295.38 kilometers per hour.

Before we compute this fraction, like before we’ll want to convert a speed in kilometers per hour to a speed in meters per second. We’ll take the same approach as before, by multiplying this fraction by 1000 meters per 3600 seconds. The product of this multiplication is 82.05 meters per second. Now we take that value and substitute it in for the 𝑣 max value given in units of kilometers per hour. Performing this division, we find that 𝑡 one is equal to 12.308 seconds. Knowing that, here is what we want to figure out. If 𝑡 one equals 12.308 seconds, then how far would Burt travel in that time?

Would it be equal to the length of the course 8.00 kilometers? If it is, then we’ve already solved for 𝑡. If not, we’ll have to take another step. Let’s refer back to our list of kinematic equations to solve for the distance that Burt will travel in a time of 𝑡 one. The third kinematic equation from the top will be useful in answering this question. It tells us that the distance Burt traveled, 𝑑, equals his initial speed times time, which is zero because he started from rest, plus one half his acceleration times time squared.

His acceleration is the acceleration we solved for earlier, 6.667 meters per second squared. And 𝑡 is 𝑡 one, 12.308 seconds. When we enter these values into this equation and multiply them together, we find a distance of 504.9 meters. This is an important result; it tells us that if we were to draw a sketch of Burt’s journey, then starting at the beginning and for some portion of that journey, but not the whole thing, his acceleration would be positive. Then for the remainder of the journey, his speed would stay constant and his acceleration would be zero.

We can call the time during which his acceleration is greater than zero 𝑡 one and the time for which his acceleration is zero 𝑡 two. We don’t yet know whether 𝑡 two is less than 𝑡 one, but we do know that if we add them together, 𝑡 one plus 𝑡 two equals 𝑡, the total time we want to solve for. Since we’ve already solved for 𝑡 one, our task is to figure out 𝑡 two. We can begin by recalling that average speed 𝑠 is equal to distance traveled divided by the time it takes to travel that distance.

In our case, 𝑣 sub max, which is equal to the remaining distance to travel which is the total course distance minus the distance we solved for all divided by 𝑡 sub two. In the numerator 8.00 kilometers is equal to 8000.0 meters. Looking at this equation, we see that we can algebraically rearrange it by switching the position of 𝑣 sub max and 𝑡 sub two.

When we do that and then enter in the value for 𝑣 sub max we solved for in meters per second, which is 82.05 meters per second, then we can perform this division and solve for 𝑡 two, which is equal to 91.35 seconds. So we now have values for 𝑡 one and 𝑡 two. And when we add them together, we get 𝑡, the time for the overall journey. 𝑡 one, 12.308 seconds, plus 𝑡 two, 91.35 seconds, gives us our total time for the journey 𝑡, which to three significant figures is 104 seconds. This is the total time it took Burt to travel the 8.00-kilometer course.