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

SS
Answered by Seeitha S. Maths tutor

3937 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

For what values of x is 2x^2 - 11x - 6 > 0 ?


Work out the value of x and y in the parallelogram ABCD.


ABC is a right-angled triangle (B being the right angle). Is AB=12 and BC=9, What is Angle A? (3.s.f.)


where do the graphs y=3x^2-2x-5 and y=4x-2 intersect


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:

© 2026 by IXL Learning