An Automobiles acceleration and speed

[swede5670]

Homework Statement

At highway speeds, a particular automobile is capable of an acceleration of about 1.6 m/s2. At this rate, how long does it take to accelerate from 83 km/h to 101 km/h?

Homework Equations

a = Change in Velocity / Change in time

The Attempt at a Solution

How do you use 1.6m/s^2 when you also have 83 km/h to 101 km/h?
I know that I need to subtract 83 from 101 (18 km/h). The car needs to increase its velocity by 18 km/h accelerating at 1.6 m/s. I don't know how to use the 1.6 m/s in the equation to show the total time

 

[Mentallic]

You need to convert the 18km/h into m/s.

18km/h = 18000m/h (convert km into m by x1000)
18000m/h = 5m/s (It travels 18000m/3600s so simply divide)
In short, to convert from km/h to m/s, just divide by 3.6
e.g. 36km/h = 10m/s


But usually with these types of questions, they are approached a little differently to develop the habit and avoid a lot of confusion later on with the more difficult questions (well that's what most of my peers do anyway)

Using the equation:
a=\frac{v-u}{t} where: v=final velocity, u=initial velocity, t=time, a=acceleration.
Note: this is the same as your equation, change in time = time, change in velocity = the final velocity - the starting velocity (101km/h - 83km/h)

Now just rearrange the formula so you make time the subject (because you are trying to find the time)

Therefore, at=v-u (multiply through by t)

t=\frac{v-u}{a} (divide through by a)

Now you have rearranged the formula so that you can plug in all the data you do know, use the calculator and end up with your answer for time.

From here you need to be sure that you convert the final and initial velocity into m/s

101km/h=\frac{101}{3.6}m/s=28\frac{1}{18}m/s

83km/h=\frac{83}{3.6}m/s=23\frac{1}{18}m/s

Plugging into the formula: t=\frac{28\frac{1}{18}-23\frac{1}{18}}{1.6}

Using the calculator: (note: should be easy to see the numerator = 5, saves time writing fractions into the calculator)
t=3\frac{1}{8} seconds

 

[swede5670]

Thanks for the help, this was really useful