Question

A person traveling at 43.2km/h applies the brake giving a deceleration of 6.0m/s2 to his scooter. How far will it travel before stopping?

Answer 1

Initial velocity of the person = 43.2 km/h = 12 m/s

Deceleration due to brakes = 6 m/s^2

Final velocity = 0 m/s

Let the distance covered be s before coming to rest

Now, using the third equation of motion:

$v^2 = u ^2 + 2as$

$0^2 = 12^ - 2\times 6 \times s$

$s = \frac{144}{12}$

s = 12 meters

Hence, the distance covered by the person is 12 meters before coming to rest.

Answer 2

According To Me,

Answer is 12 m

Given: v=43.2 km/h(see acceleration in m/s² so change it to m/s)

43.2*5/18=12 m/s<----velocity in m/s

Acceleration.=6 m/s²

You see the person at this time travel with initial velocity(u)=12 and after sometime he stop me means final velocity(v)=0

You Find Distance from[--S=ut+1/2at²--]  S=distance ,t=time ,u=initial velocity----1

You have u=12 ,a=-6(minus becz its accl decrease) but you don't know (t)

So find t

v=u + at (v=0,u=12,a=-6,t=find)-------2

0=12 - 6t after solving t=2sec

So you have t=2sec put in -----1

S=12 * 2 - 1/2 * 6 * 2²

S=24-12

S=12m----Answer

I hope you understand the answer

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