Question

A driver of a car moving at 30 m/s suddenly notices a child 80m straight ahead. If the driver’s reaction time is 0.5 s and the deceleration is 8 m s -2 , can he avoid hitting the child?

Answer 1

Given:

A driver of a car moving at 30 m/s suddenly notices a child 80m straight ahead. If the driver’s reaction time is 0.5 s and the deceleration is 8 m/s².

To find:

If the driver can avoid hitting the child.

Calculation:

Reaction time of driver is 0.5 seconds.

So, distance between car and child after the reaction time be d ;

$d = 80 - (30 \times 0.5)$

$= > d = 80 - 15$

$= > d = 65 \: m$

Now , deceleration is 8 m/s².

So , let the distance taken to stop be x ;

${v}^{2} = {u}^{2} + 2ax$

$= > {0}^{2} = {(30)}^{2} + 2( - 8)x$

$= > 0 = 900 - 16x$

$= > 16x= 900$

$= > x = 56.25 \: m$

Since x < d , so the the driver can avoid hitting the child.

So, final answer:

Child will not be hit by the car.

Answer 2

Answer:

Given:

A driver of a car moving at 30 m/s suddenly notices a child 80m straight ahead. If the driver’s reaction time is 0.5 s and the deceleration is 8 m/s².

To find:

If the driver can avoid hitting the child.

Calculation:

Reaction time of driver is 0.5 seconds.

So, distance between car and child after the reaction time be d ;

d = 80 - (30 \times 0.5)d=80−(30×0.5)

= > d = 80 - 15=>d=80−15

= > d = 65 \: m=>d=65m

Now , deceleration is 8 m/s².

So , let the distance taken to stop be x ;

{v}^{2} = {u}^{2} + 2axv

2

=u

2

+2ax

= > {0}^{2} = {(30)}^{2} + 2( - 8)x=>0

2

=(30)

2

+2(−8)x

= > 0 = 900 - 16x=>0=900−16x

= > 16x= 900=>16x=900

= > x = 56.25 \: m=>x=56.25m

Since x < d , so the the driver can avoid hitting the child.

So, final answer:

Child will not be hit by the car.