Question

Question #73ff4

Answer 1

`11.53s`

Explanation:

`u->"Initial velocity"=26m/s`

`v->"Final velocity"=0`

`s->"Distance covered"=150m`

`"Retardation required"=am/s^2`

`v^2=u^2-2*a*s`

`=>0^2=26^2-2*a*150`

`=>a=26^2/300=2.253m/s^2`

But car has got the capacity to apply maximum retardation `2.5m/s^2`.So accident can be avoided.

To calculate the time t available to avoid accident applying uniform retardation as calculated above we can use the equation

`s=(u+v)/2*t`

`=>150=(26+0)/2*t`

`t=150/13~~11.53s`
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Alternate solution with different assumptions.

This alternate solution assumes

1. Driver can delay application of brakes after seeing the child.
2. The car comes to dead stop in `150m` distance after seeing the child.
3. When brakes are applied the given retardation is produced.

Suppose driver applies brakes immediately after seeing the child ahead.

Using the kinematic equation
`v=u+at`
where `v,u,a and t` are final velocity, initial velocity, acceleration and time taken respectively.

Inserting given numbers we get
`0=26-2.5t`
`=>t=26/2.5=10.4s`

`:.` Distance `s` traveled in these `10.4s` is given by
`v^2-u^2=2as`
`0^2-(26)^2=2xx(-2.5)s`
`=>s=(26)^2/5=135.2m`.
We see that in this case child is always safe. As already evident from the first solution.

Suppose the driver delays application of brakes after seeing the child.
Distance which the car can travel at maximum speed before application of brakes`=150-135.2=14.8m`
Time taken to cover this distance`="Distance"/"Speed"=14.8/26`
`=0.57s`, rounded to two decimal places.

Time available to avoid hitting the child in this case`=10.4+0.57=10.97s`, rounded to two decimal places.