physics homework question

Question

An object moves along the x-axis according to the equation x = 2.95t² − 2.00t + 3.00, where x is in meters and t is in seconds.

a- Determine the instantaneous speed at t = 4.20 s.

b-Determine the instantaneous acceleration at t = 4.20 s.

Answer 1

x =2.95t^2 -2t +3

speed = x' = 5.9t-2

x'(4.2)= 5.9(4.2)-2= 24.78-2= 22.78 m/s

acceleration = x"= 5.9 m/s^2

Answer 2

Raymond's answer is correct, however it requires calculus. Seeing that Algebra 1 is tagged in your question I'm assuming this is algebra-based physics with very little calculus. Here is an alternative method to find the answer.

If we assume a constant acceleration - a reasonable assumption seeing that position function is parabolic - then we can use the constant acceleration equation to solve for position and relate that to the given position function.

We know that when acceleration is constant:

Δx=v₀t+(1/2)at²

Given the position equation, x=2.95t² − 2.00t + 3.00, we can move the +3.00 to the other side to find the equation for Δx at a time t:

( x-3.00)=Δx=2.95t² − 2.00t + 3.00

comparing the above equation with the constant acceleration equation we know that (1/2)a=2.95 and v₀=-2.00.

now we can solve for the acceleration and find that a=5.90 m/s^2. Finally to find the instantaneous velocity at t=4.20s we can just use the acceleration we found and plug that into the equation v=v₀+at with v₀=-2.00, a=5.90 m/s^2, and t=4.20s. This gives you the final answer v=22.78 m/s.