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

3949 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

How do you find the highest common factor of two numbers?


A 4 digit number is picked. It's second digit is a prime number, it must be even and it must be greater than 5000. How many possible numbers can be picked?


Which of these lines are parallel to y=2x+3? Which are perpendicular? Options: 1) y=5x-4, 2) y=-1/3x+3, 3) y=-1/2x-1, 4) y=2x-2/3


Equations


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