Question

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qus b​

Answer 1

According to the question coefficient of friction=0.5

normal reaction=mg

frictional force=0.5mg

let applied force=ma

resultant force in horizontal direction =m(a-0.5g)

acceleration=a-0.5g

now , v=u+at

⇒72=(a-5)1hr

⇒a=77km/hr²

again s=ut+1/2at²

⇒distance travelled by the car=77÷2=39.5 km

⇒