Question

A cyclist traveling at 7m / s sees with horror an automobile door opening in front of him, at a distance of 11m. Its reflex time (to apply the brakes and see the door) is 0.400s.a) What must be its acceleration to just avoid the collision.b) How long does it take to cross the distance of 11 m?

Answer 1

Given data

*The speed of the cyclist is u = 7 m/s

*The given distance is s = 11 m

*The given time is T = 0.400 s

(a)

The formula for the acceleration of the cyclist to just avoid the collision is given by the equation of motion as

`\begin{gathered} v^2=u^2+2as \\ a=(v^2-u^2)/(2s) \end{gathered}`

*Here v = 0 m/s is the final speed of the cyclist

Substitute the known values in the above expression as

`\begin{gathered} a=((0)^2-(7)^2)/(2*11) \\ =-2.22m/s^2 \end{gathered}`

Hence, the acceleration of the cyclist to avoid the collision is a = -2.22 m/s^2

(b)

The formula for the time taken to cross the distance of 11 m is given by the equation of motion

`s=ut+(1)/(2)at^2`

Substitute the known values in the above expression as