The gradient on a distance-time graph gives the velocity, therefore if an equation involves distance and time then it can be differentiated to give velocity.
Similarly, the gradient on a velocity-time graph gives acceleration. Differentiating a velocity-time graph will give acceleration.
A particle is moving between two nodes. The displacement of the particle is given by s=t3-2t2-12t. What is the velocity of the particle after 5 seconds?
Displacement | s | =t3-2t2-12t |
dsdt = velocity | dsdt | =3t2-4t-12 |
v | =3t2-4t-12 | |
At 5 seconds | =3(5)2-4(5)-12 | |
=43ms-1 |
Answer: 43ms-1
The displacement of a particle is given by s=t3-t2-t. Determine when the velocity will be zero.
Displacement | s | =t3-t2-t |
velocity = | dsdt | =3t2-2t-1 |
When is velocity zero? | 0 | =3t2-4t-1 |
Use solution for quadratic | x | =-b±√b2-4ac2a |
x | =-(-2)±√(-2)2-4(3)(-1)2(3) | |
x | =1 |
Ignore the negative time value, as it makes no sense in this context.
Answer: After 1 second