## Answers

Since, Force required to stop is given by,

F = m*a

here, m = mass of rider = 99 kg

a = de-acceleration = 1.8 m/sec^2

So, F = 99*1.8

**F = 178.2 N**

**b.)**

by kinematics law,

v^2 = u^2 + 2*a*S

here, v = final velocity = 0 m/sec.

u = initial velocity = 34 m/sec.

a = -1.8 m/sec.^2

S = (v^2 - u^2)/(2*a) = (0^2 - 34^2)/(2*(-1.8))

**S = 321.11 m**

**c.)**

Work done by force = W = F*S*cos(x)

here, x = angle between force and distance vector = 180 deg

So, W = 178.2*321.11*cos(180 deg)

**W = -5.72*10^4 J**

d.)

Now, a = 3.6 m/sec^2

F = 99*3.6

**F = 356.4 N**

e.) Since, distance travel is given by ,

S = (v^2 - u^2)/(2*a) = (0^2 - 34^2)/(2*(-3.6))

**S = 160.56 m**

f.)

also, W = 356.4*160.56*cos(180 deg)

**W = -5.9*10^4 J**

Please upvote.

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