Edexcel IGCSE Math A 3.4.HE calculus in kinematics

Practise kinematics-calculus questions by differentiating displacement to get velocity and differentiating again for acceleration.

Exam points

Differentiate s(t) to find velocity at a given time or when a condition is met.
Differentiate velocity to find acceleration and interpret the time value in context.
Solve equations from velocity or acceleration conditions using the allowed time range.

[Maximum number: 4]

A particle is moving along a straight line that passes through the fixed point O The displacement, s metres, of the particle from O at time t seconds is given by

s=2 t^{3}-5 t^{2}+6 t-5

Find the value of t when the acceleration of the particle is $5 \mathrm{~m} / \mathrm{s}^{2}$

s=2t^3-5t^2+6t-5

]

v=\frac{ds}{dt}=6t^2-10t+6

]

a=\frac{dv}{dt}=12t-10

]

12t-10=5

]

12t=15

]

Final answer:

t=1.25

]

M1 for $\frac{ds}{dt}=6t^2-10t+6$ , with at least 2 terms correct.
M1ft for $\frac{dv}{dt}=12t-10$ , following from a 3-term quadratic.
M1ft for setting 12t-10=5, dependent on the previous M1.
A1 oe for 1.25.

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3.4.HE Calculus in kinematics and practical problems question 1