M!
the velocity v at time t of an object travelling in a straight line is given by v = t - 3.  determine the distance it travels from t = 0 and t = 4.

5

Step-by-step explanation:

To solve with calculus, distance is the integral of speed:

d = ∫ |v| dt

d = ∫₀⁴ |t − 3| dt

d = -∫₀³ (t − 3) dt + ∫₃⁴ (t − 3) dt

d = ∫₀³ (3 − t) dt + ∫₃⁴ (t − 3) dt

d = (3t − ½ t²) |₀³ + (½ t² − 3t) |₃⁴

d = [ (9 − 9/2 ) − (0 − 0) ] + [ (8 − 12) − (9/2 − 9) ]

d = 9/2 + 1/2

d = 5

You can also find this geometrically.  Graph y = |x − 3|, then find the area under the curve.  You will find it's the area of two triangles.

d = ½ (3)(3) + ½ (1)(1)

d = 5

It's important to note that distance is not the same thing as displacement.  Displacement is the difference between where you start and where you stop.  Distance is length of the path you take.

may i see the picture? it would a lot, xox

u = -(6/7)w

step-by-step explanation:

place all values with 'u' on one side of the equation.

4u + 8w = - 3u + 2w

4u + 3u = 2w - 8w

further simplify the equation by solving the addition and subtraction.

7u = - 6w

make 'u' the independent variable by placing its co-efficient on the right-hand side.

u = - (6/7)w