Velocity and acceleration from a position function
Differentiate a position function to find velocity, and differentiate again for acceleration. Unlimited questions, five difficulty levels, and a full worked solution every time — free.
Try one
The height, in metres, of a toy rocket t seconds after launch is s(t) = -2t^3 - 2t^2 - 2t. Find the rocket's acceleration at t = 1 seconds.
s(t) = −2t3 - 2t2 - 2t
Answer: -16
See one solved, step by step
The height, in metres, of a toy rocket t seconds after launch is s(t) = 3t^3 + t^2 - 3t. Find the rocket's acceleration at t = 1 seconds.
s(t) = 3t3 + t2 - 3t
📘 Worked solution
1v(t) = s'(t) = 9t2 + 2t - 3Differentiate the position function to get velocity.
2a(t) = v'(t) = 18t + 2Differentiate the velocity function to get acceleration.
3a(1) = 18(1) + 2 = 20Substitute t = 1.
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