Question

somebody explain fourth question.​

Answer 1

Answer:

(4) - 8 m/s

Explanation:

We will approach the problem in this way,

1. We will find equation of parabola.
2. We will differentiate this curve with respect to time at 3 seconds.
3. ds/dt would be velocity.

The parabola looks like

(x-h)² = -4a (y-k)

From inference it is clear that curve passes through (2.5,50); (0,0); (5,0).

Substitute this points in above equation one by one,

1) (2.5-h)² = -4a (50 -k)

2) h² = 4ak

3) (5-h)² = 4ak

From 2) and 3)

5-h = h

h = 2.5

Substitute this in 1)

0 = -4a(50 - k)

a cannot be zero for a parabola,

so k = 50

Substitute values of k and h in 2)

(2.5)² = 4 a 50

2.5 x 2.5 = a x 4 x50

a = 0.03125.

Therefore the parabola is

(x-2.5)² = -4(0.03125)(y-50)

(x-2.5)² = 0.125(50-y).

But x axis is time and y axis is distance.

(t-2.5)² = 0.125(50-s)

Differentiate this equation with respect to t

Math help: d/dx (xⁿ) = n x ⁿ⁻¹.

2(t-2.5) = 0.125(-ds/dt)

ds/dt = 2 (2.5-t)/0.125

v = 16(2.5-t)

We need to find velocity at 3 s so t = 3

∴v = 16(2.5 - 3)

v = 16(-0.5)

v = -8 m/s.