An object's displacement, s metres, from a fixed point after t seconds is s=5t^3+t^2. Find expressions for the object's velocity and acceleration at time t seconds.

Differentiating gives the rate of change and velocity and acceleration are rates of change. Velocity is the rate of change of displacement compared to time and acceleration is the rate of change of velocity compared to time. Therefore, differentiating an expression for displacement in terms of time, gives velocity and differentiating an expression for velocity in terms of time, gives acceleration.Therefore, v=15t^2+2t and a=30t+2

Answered by Seeitha S. Maths tutor

2970 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

what is the median, mode and mean?


Heather invests £900 for 3 years at 2% per annum compound interest. Calculate the value of her investment at the end of the 3 years.


v^2 = 2w - x^2. w = 40; x = 4. Find the value of v.


ln(2x^2 +9x-5) =1+ ln( x^2+2x-15)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences