What is d^2s/dt^2 equivalent to?
Differentiating displacement (s) twice so acceleration
The displacement is given by: s=2t^3 -6t. Find expressions for the velocity and acceleration of the object at time ‘t’
Velocity: (differentiate once) = 6t^2 -6
Acceleration: (differentiate twice) = 12t
The velocity of a model train is v=0.3t^2 -0.5. Find its displacement from its initial position after 3 seconds
1) Integrate velocity to get displacement: s= ∫(0.3t^2 -0.5)dt =0.1t^3 - 0.5t +c
2) ‘Initial position’ means you can use s=0 and t=0 to find c: 0=0.1(0)^3 - 0.5(0) +c so c=0
3) So the equation is now: s=0.1t^3 -0.5t
4) Substitute in t=3: s=0.1(3)^3 -0.5(3) =1.2 m
What does it mean when a variable acceleration question states ‘initial conditions’?
You can assume that v=0, s=0, a=0 and t=0 unless otherwise stated
A car moves between 2 sets of traffic lights, stopping at both. Its speed is modelled by: v= t/20(40-t). Find the times at which the car is stationary and the distance between the 2 sets of traffic lights.
1) the car is stationary when v=0 so find the roots of the quadratic: t/20(40-t) gives roots of zero and 40
2) the distance between the traffic lights is the area of the velocity-time graph (displacement) so integrate v to find s: ∫1/20(40-t) with bounds of 40 and zero gives you 533 metres
a= k-18t^2. At time t=0 the particle passes through the origin with velocity 7ms^-1. One second later the particle passes through point Q with velocity 9ms^-1. Find:
1) the value of k
2) the distance of Q from the origin
1) v=kt-6t^3 +c, when t=0 & v=0 c=7 so v=kt-6t+7.
When v=9, t=1 so 9=k-6+7 thus k=8
2) s=4t^2 -3/2t^4 + 7t + c, when s=0 & t=0 c=0 so s= 4t^2 -3/2t^4 + 7t. Substitute in zero: 0m
Substitute in 1: 4-3/2+7=9.5m
9.5-0=9.5m
A particle starts from rest at a point A at time t=0, where t is in seconds. The particle moves in a straight line. For 0≤t≤4 the acceleration is 1.8t ms^-2 and for 4≤t≤7 the particle has constant acceleration 7.2ms^-2. Find an expression for the velocity of the particle in terms of t, valid for 0≤t≤4
v= ∫a so v= ∫1.8t = 0.9t^2 + c
When v=0, t=0 so c=0
Thus v=0.9t^2
