We are always trying to put the equation in to the form:

y = mx + c

 

 

a) p = mv

p/v = m

p(1/v) = m

m = p(1/v)

So to get a straight line graph we will plot m against 1/v

Gradient will be p.

c = 0
Therefore intercept will be (0,0)

b) T = 2π √(L/g)

So to get a straight line graph we will plot T² against L

Gradient will be 4π²/g.
Thus g can be found from g = 4π²/(gradient)

c = 0
Therefore intercept will be (0,0)

c) s = ½at²

s = ½at²

So to get a straight line graph we will plot s against t²

Gradient will be ½a
Thus a can be found from a = 2 × (gradient)

c = 0

d) n = (sin i) / (sin r)

(sin r)n = (sin i)

(sin i) = n(sin r)

So to get a straight line graph we will plot (sin i) against (sin r).

Gradient will be n.

c = 0
Therefore intercept will be (0,0).

e) 1/ƒ = 1/u + 1/v

(1/u) = (–1)(1/v) + (1/ƒ)

(1/u) = (–1)(1/v) + (1/ƒ)

So to get a straight line graph we will plot 1/u against 1/v

Gradient will be -1.

c = (1/ƒ)
Thus intercept will be (0,(1/ƒ))

ƒ can be found from ƒ = 1/(gradient)