Find instantaneous acceleration of a powered vehicle on a hill
Combine P/v with resistance and the slope weight component, distinguish instantaneous acceleration from cruise speed and avoid a constant-acceleration assumption.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.5. Use g = 10 m s⁻² throughout these problems. Masses are positive and constant. Work by a constant force is Fd cos θ, with θ between force and displacement; scalar-product notation is not required. Kinetic energy is ½mv² and gravitational potential-energy change is mgΔh. State the system and count gravitational work or potential-energy change once, not both. Mechanical energy is conserved only under the stated loss-free model; friction and supplied resistance transfer mechanical energy to other forms. Smooth curved-slide examples use overall energy changes and assume contact throughout, rather than requiring a normal-reaction calculation. Average power is work divided by elapsed time; P = Fv applies to a force along the motion. Constant driving power generally gives variable force and acceleration; instantaneous calculations do not justify constant-acceleration formulae over a whole journey. Check string, contact and motion-state changes before continuing a model. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Use F = ma and the along-slope gravitational component.
- Use instantaneous P = Fv and convert kilowatts to watts.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A 40 kW drive acts at uphill speed 20 m s⁻¹. What is its driving force?
What you will learn
- Find driving force and instantaneous acceleration at a specified positive uphill speed.
- Calculate a possible steady speed and interpret speeding up or slowing down relative to it.
Power supplies a drive force at the current speed
A 1000 kg vehicle moves uphill at 20 m s⁻¹ with driving power 40 kW, resistance 500 N and sin θ = 1/20. Its drive force at this instant is 40 000/20 = 2000 N. The downslope weight component is 1000 × 10 × 1/20 = 500 N. With up the hill positive, 2000 − 500 − 500 = 1000a, so a = +1 m s⁻². The normal reaction does no along-path work under the fixed-road model.
At a larger speed with the same power and constant resistance, driving force is smaller and the instantaneous acceleration changes. At speed 50, the drive is 800 N while total opposition remains 1000 N, giving a = −0.2 m s⁻². The vehicle still moves uphill but slows. The acceleration sign does not immediately reverse its current direction of travel.
A steady speed is a force balance, not a whole-journey average
For this example, zero acceleration requires P/v = 500 + 500 = 1000 N, so possible steady speed is v = 40 000/1000 = 40 m s⁻¹. Speeds below 40 have positive acceleration and speeds above 40 have negative acceleration while the same model holds. The value is a model prediction, not permission to infer travel time or a real safe road speed.
Constant power does not make acceleration constant over an interval because P/v changes with v. Do not substitute the acceleration at 20 into v = u + at for an entire speed-changing journey. A downhill problem needs its own signed gravitational component, and a speed-dependent resistance must be evaluated from its supplied law. Our numerical examples use specified positive uphill speeds and constant opposition; a zero or negative denominator needs a different interpretation, not an automatic finite cruise speed.
Pause and explain
In the 1000 kg hill example, speed 50 m s⁻¹ gives drive 800 N and total opposition 1000 N. What follows?
Worked example
A 1000 kg vehicle moves uphill at 20 m s⁻¹, with power 40 kW, resistance 500 N and sin θ = 1/20. Find its instantaneous acceleration and possible steady speed in this model.
Show the worked solution
- At the stated speed, drive force is 40 000/20 = 2000 N and downslope weight component is 500 N.
- Newton’s law gives a = (2000 − 500 − 500)/1000 = 1 m s⁻² uphill.
- For zero acceleration, v = 40 000/(500 + 500) = 40 m s⁻¹, provided the model remains valid.
Answer Instantaneous a = +1 m s⁻²; possible steady speed 40 m s⁻¹.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A 1000 kg vehicle moves uphill at 20 m s⁻¹ with drive power 40 kW, resistance 500 N and sin θ = 1/20. Find its instantaneous acceleration.
Give me a hint
Calculate the drive from the current speed and include both opposing forces.
Compare my reasoning
- Drive force is 40 000/20 = 2000 N.
- The slope weight component is 1000 × 10 × 1/20 = 500 N, so resultant is 2000 − 500 − 500 = 1000 N uphill.
- Acceleration is 1000/1000 = +1 m s⁻² at this instant.
Instantaneous a = +1 m s⁻² uphill.
Look for these in your work
- Convert kilowatts and use the instantaneous speed.
- Subtract resistance and the slope weight component separately.
A 1200 kg vehicle moves uphill at 15 m s⁻¹ with power 30 kW, resistance 400 N and sin θ = 1/20. Find its instantaneous acceleration and possible steady speed under this constant-resistance model.
Give me a hint
The steady-speed condition sets resultant force to zero at a potentially different speed.
Compare my reasoning
- At 15, drive is 30 000/15 = 2000 N and slope weight component is 1200 × 10/20 = 600 N.
- Instantaneous acceleration is (2000 − 400 − 600)/1200 = 5/6 m s⁻² uphill.
- For zero acceleration, drive equals 1000 N, so possible steady speed is 30 000/1000 = 30 m s⁻¹.
Instantaneous a = 5/6 m s⁻²; possible steady speed 30 m s⁻¹.
Look for these in your work
- Use total opposing force in both calculations.
- Do not equate the stated current speed with the predicted steady speed.
The 1000 kg vehicle with constant power 40 kW, resistance 500 N and sin θ = 1/20 is moving uphill at 50 m s⁻¹. Find its instantaneous acceleration, explain whether it is speeding up and decide whether that acceleration can be used for the whole later journey.
Give me a hint
Compare current drive P/v with the unchanged total opposing force.
Compare my reasoning
- Current drive is 40 000/50 = 800 N while total opposition is 1000 N.
- Thus a = (800 − 1000)/1000 = −0.2 m s⁻²; its positive uphill velocity is decreasing in magnitude.
- As speed changes, P/v and acceleration change, so −0.2 is an instantaneous value and cannot be assumed constant over the whole later journey.
Instantaneous a = −0.2 m s⁻²; uphill speed decreases; whole-journey constant acceleration is not justified.
Look for these in your work
- Keep the current velocity direction distinct from acceleration direction.
- State the instantaneous limitation of the constant-power force calculation.
Common mistakes
- Subtract both resistance and the uphill gravitational opposition.
- An instantaneous acceleration is not constant over a whole constant-power journey.
- A negative acceleration can mean slowing while still moving uphill.
What can you explain now?
May the instantaneous acceleration at one speed be used as constant throughout a speed-changing constant-power journey?
Compare with the explanation
No; the driving force P/v changes with speed, so acceleration generally changes.
Use a separately justified interval model if calculating elapsed travel time.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
At the specified positive speed, use P/v for drive, subtract all signed opposition and divide the resultant by mass.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.