jmaths
IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.21
Name Date
Kinematics on a line Problem set A
Show the derivative or integral you use. Time is in seconds, displacement in metres, and your calculator is in radians.
Difficulty: first skills applying them stretch
Skills
-
The displacement \(s\) metres of an object from a fixed point at time \(t\) seconds is given. Find expressions for the velocity and acceleration.
- (a) \(s = 4\cos 5t\)
- (b) \(s = 3e^{-2t}\)
- (c) \(s = (2t + 1)^{4}\)
-
A particle moves on a line. Its velocity \(v\) is given in terms of its displacement \(s\). Find an expression for the acceleration in terms of \(s\).
- (a) \(v = 3s + 2\)
- (b) \(v = 4 - s^{2}\)
- (c) \(v = e^{s}\)
-
Use your calculator to evaluate each definite integral. Give each answer to 3 significant figures. Time is in seconds.
- (a) \(\displaystyle\int_1^3 5\sin 2t\,dt\) (change in velocity, \(a = 5\sin 2t\))
- (b) \(\displaystyle\int_0^2 3te^{-t}\,dt\) (change in displacement, \(v = 3te^{-t}\))
In context
-
A piston rod moves in a straight line. Its displacement from the centre of the cylinder is \(s = 0.04\sin 50t\) metres at time \(t\) seconds.
- (a) Find the velocity when \(t = 0.2\).
- (b) Find the acceleration when \(t = 0.2\).
-
A cyclist brakes as she approaches a junction. Her velocity is \(v = 24 - 3s\) m s⁻¹ when she is \(s\) metres past the point where she starts braking.
- (a) Find an expression for the acceleration in terms of \(s\).
- (b) Find the acceleration when \(s = 4\). What does the sign tell you?
-
A rocket sled's acceleration is \(a = 9e^{-0.3t}\) m s⁻², with \(t\) in seconds.
- (a) Find the change in velocity between \(t = 2\) and \(t = 6\).
- (b) The velocity is 15 m s⁻¹ when \(t = 2\). Find the velocity when \(t = 6\).
Stretch
-
A particle moves on a line with displacement \(s = 5\sin 2t + 1\) metres.
- (a) Find an expression for \(a\) in terms of \(t\).
- (b) Find the constant \(k\) such that \(a = k(s - 1)\).
-
A particle moves on a line. Its velocity satisfies \(v^2 = 16 - s^2\). Use \(a = \dfrac{d}{ds}\left(\tfrac12 v^2\right)\) to find an expression for the acceleration in terms of \(s\).
jmaths.xyz · AI HL · AHL 4.21 Kinematics on a line: derivatives and integrals