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PhysicsPrimary schoolEasy

Distance in uniformly accelerated motion

A car starts from rest and moves in uniformly accelerated motion with acceleration a=2 m/s2a = 2\ \text{m/s}^2.

  1. How far does it travel in the first 5 s5\ \text{s}?
  2. What speed does it reach after that time? Give it in m/s\text{m/s} and in km/h\text{km/h}.
Hintshow

The car starts from rest, so its initial speed is zero. Then s=at22s = \dfrac{a t^2}{2} and v=atv = a t.

Solutiontry it on your own first

Step 1. Write down what we know and what we want

The car starts from rest, so its initial speed is zero:

v0=0a=2 m/s2t=5 s\begin{aligned} v_0 &= 0 \\ a &= 2\ \text{m/s}^2 \\ t &= 5\ \text{s} \end{aligned}

We want the distance ss and the final speed vv.

Step 2. Choose the formula for the distance

For uniformly accelerated motion the distance is:

s=v0t+at22s = v_0 t + \frac{a t^2}{2}

Because v0=0v_0 = 0, the first term is zero:

s=at22s = \frac{a t^2}{2}

Step 3. Work out the distance

Substitute the values:

s=2 m/s2⋅(5 s)22s = \frac{2\ \text{m/s}^2 \cdot (5\ \text{s})^2}{2}

First square the time on its own:

(5 s)2=25 s2(5\ \text{s})^2 = 25\ \text{s}^2

Multiply by the acceleration - the seconds squared cancel and metres are left:

2 m/s2⋅25 s2=50 m2\ \text{m/s}^2 \cdot 25\ \text{s}^2 = 50\ \text{m}

Divide by 2:

s=50 m2=25 m\boxed{s = \frac{50\ \text{m}}{2} = 25\ \text{m}}

Step 4. Work out the speed

In uniformly accelerated motion the speed goes up by aa every second:

v=v0+atv = v_0 + a t

Substitute v0=0v_0 = 0:

v=0+2 m/s2⋅5 sv = 0 + 2\ \text{m/s}^2 \cdot 5\ \text{s}
v=10 m/s\boxed{v = 10\ \text{m/s}}

Step 5. Convert m/s to km/h

An hour has 36003600 seconds and a kilometre has 10001000 metres. So:

1 ms=3600 m1 h=3.6 kmh1\ \frac{\text{m}}{\text{s}} = \frac{3600\ \text{m}}{1\ \text{h}} = 3.6\ \frac{\text{km}}{\text{h}}

Multiply our speed by 3.63.6:

v=10⋅3.6 km/h=36 km/hv = 10 \cdot 3.6\ \text{km/h} = 36\ \text{km/h}

Step 6. Check the result

With constant acceleration, the average speed vˉ\bar{v} is the average of the initial and final speeds:

vˉ=0+10 m/s2=5 m/s\bar{v} = \frac{0 + 10\ \text{m/s}}{2} = 5\ \text{m/s}

Over 5 s5\ \text{s} that gives 5 m/s⋅5 s=25 m5\ \text{m/s} \cdot 5\ \text{s} = 25\ \text{m} - the same as in step 3.

Answer

s=25 m\boxed{s = 25\ \text{m}}
v=10 m/s=36 km/h\boxed{v = 10\ \text{m/s} = 36\ \text{km/h}}

In the first 5 seconds the car travels 25 metres and reaches 10 m/s, which is 36 km/h.

Sources

We checked the facts on this page against these sources. All of them may be reused freely: they carry an open licence or are not covered by copyright.

  1. Martin, Neary, Rinaldo and Woodman, Introductory Physics: Building Models to Describe Our World (LibreTexts) - 3.2 Motion with Constant Acceleration (x = x0 + v0t + at²/2, v = v0 + at) · CC BY-SA 4.0
  2. Wikibooks - FHSST Physics/Rectilinear Motion/Equations of Motion (v = u + at, s = ut + at²/2, a car starting from rest) · CC BY-SA 4.0
  3. Jeffrey W. Schnick, Calculus-Based Physics (LibreTexts) - 7A One-Dimensional Motion: The Constant Acceleration Equations · CC BY-SA 2.5
  4. BIPM - The International System of Units (SI Brochure, 9th edition), Table 8: 1 h = 3600 s, and the example 25 m/s = 90 km/h (PDF) · CC BY 4.0
  5. NIST - Guide to the SI (SP 811), Appendix B.9: conversion factors (kilometer per hour to meter per second) · public domain