Rectilinear Kinematics
Time & Distance to Reach Target Speed — Constant Acceleration
A car travels along a straight road with an initial speed of 70 km/h
and a constant acceleration of 6000 km/h².
Determine how long it takes the car to reach a speed of 120 km/h,
and the distance traveled during this acceleration phase.
What We Know
- Initial speed: v₀ = 70 km/h
- Final speed: v = 120 km/h
- Constant acceleration: a = 6000 km/h²
- Find: time t to reach 120 km/h
- Find: distance s traveled during this time
Why constant-acceleration formulas? The acceleration is given as a fixed value — not a function of time or position. This activates the standard kinematic equations. All quantities are already in consistent units (km, h), so no conversion is needed before solving. However, converting the final answers into seconds and metres makes them more physically intuitive.
Part A — Time to Reach 120 km/h
Step 1
Apply v = v₀ + at and Solve for t
The first kinematic equation links velocity, initial velocity, acceleration, and time directly:
\[ v = v_0 + at \]
Substituting the known values:
\[ 120 = 70 + (6000)\,t \]
\[ 6000\,t = 120 – 70 = 50 \]
\[ t = \frac{50}{6000} = \frac{1}{120} \ \text{h} \]
\[ t = 0.00833 \ \text{h} \]
In hours0.00833 h
In seconds30 s
Unit check & intuition: t = 1/120 h × 3600 s/h = 30 s. It takes exactly half a minute to accelerate from 70 to 120 km/h — a very aggressive but realistic acceleration (6000 km/h² ≈ 1.54 m/s², comparable to a brisk highway merge).
Part B — Distance Traveled
Step 2
Apply v² = v₀² + 2as and Solve for s
The position–velocity kinematic equation is ideal here — it gives distance directly without needing the time result:
\[ v^2 = v_0^2 + 2as \]
Substituting known values:
\[ (120)^2 = (70)^2 + 2(6000)\,s \]
\[ 14400 = 4900 + 12000\,s \]
\[ 12000\,s = 14400 – 4900 = 9500 \]
\[ s = \frac{9500}{12000} = \frac{19}{24} \ \text{km} \]
\[ s = 0.792 \ \text{km} \]
In kilometres0.792 km
In metres792 m
Cross-check using s = v₀t + ½at²:
s = 70(1/120) + ½(6000)(1/120)²
= 70/120 + 3000/14400
= 0.5833 + 0.2083 = 0.7917 km ✓ — matches perfectly.
Step 3
Summary of Results in Both Unit Systems
| Quantity | Value (km–h system) | Value (m–s system) |
|---|---|---|
| Initial speed v₀ | 70 km/h | 19.44 m/s |
| Final speed v | 120 km/h | 33.33 m/s |
| Acceleration a | 6000 km/h² | 0.463 m/s² |
| Time t | 0.00833 h | 30 s |
| Distance s | 0.792 km | 792 m |
Why keep km/h units throughout? All three given quantities share the same unit system (km and h), so working in those units avoids conversion errors mid-calculation. Converting only the final answers to SI (m and s) gives the most physically meaningful presentation — 30 s and 792 m are far easier to visualise than 0.00833 h and 0.792 km.
Final Answers
TIME TO REACH 120 km/h
t = 0.00833 h = 30 s
DISTANCE TRAVELED
s = 0.792 km = 792 m