A car is moving on an inclined road with friction acting upon it. When it is moving up the road at a speed v the engine is working at power 3P and when it is moving down the road at v the engine is working at a power P. Find the value of P.

Incline is at θ where sin θ = 1/20 and mass of the car is 800kg and v is 12.5 m/s

Up the road: Power = Fv                              F = R + (800g)/20

                                             Power = (R + 40g)*25/2 = 3P …. P = (R + 40g)*25/6

Down the road: Power = Fv          F = R – (800g)/20

                                             Power = (R - 40g)*25/2 = P

Equate the equations:

(R + 40g)*25/6 = (R – 40g)*25/2

25R + 1000g = 75R – 3000g so R = 80g

Therefore P = (80g – 40g)*25/2 = 500g = 4900W

JM
Answered by Jonathan M. Maths tutor

3078 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Differentiate: y = xsin(x)


A spherical balloon of radius r cm has volume Vcm^3 , where V =4/3 * pi * r^3. The balloon is inflated at a constant rate of 10 cm^3 s^-1 . Find the rate of increase of r when r = 8.


Find the coordinates of the point of intersection between the line L:(-i+j-5k)+v(i+j+2k) and the plane π: r.(i+2j+3k)=4.


Show that Sec2A - Tan2A = (CosA-SinA)/(CosA+SinA)


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning