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A car starting from rest moves with a constant acceleration of 10 mi/hr2 for 1 hour, then decelerates at a constant 5 mi/hr2 until it comes to a stop. How far has it travelled?

Answer :

Answer:

The total distance traveled by the car (S) = 15 mi

Explanation:

Total acceleration ([tex]a_{1}[/tex]) = 10 mi / [tex]h^{2}[/tex]

Time (t) = 1 hour

Deceleration ([tex]a_{2}[/tex]) = - 5 mi / [tex]h^{2}[/tex]

From law of speed we know that  V = u + [tex]a_{1}[/tex] t -----------(1)

Where V = Final speed

           u = initial speed and t = time period

Car starts from rest so initial speed is zero so u = 0

Then equation 1 becomes V = [tex]a_{1}[/tex] t = 10 × 1 = 10 mi / h ---------(2)

The distance traveled by the car [tex]S_{1}[/tex] = u t + [tex]\frac{1}{2}[/tex] × [tex]a_{1}[/tex] × [tex]t^{2}[/tex] -------------(3)

Put all the values in equation we get [tex]S_{1}[/tex] = 0 + [tex]\frac{1}{2}[/tex] × 10 × [tex]1^{2}[/tex]

                                                             [tex]S_{1}[/tex] = 5 mi ---------------(4)

This is the distance traveled by the car in 1 hour.

In second case when car starts decelerates the equation of motion becomes

⇒ V = u - a t -----------(5)

final speed in this case = 0 and initial speed (u) = 10 mi / h

⇒ 0 = 10 - 5 t

⇒ 5 t = 10

⇒ t = 2 hours

Distance traveled by the car in to hours  [tex]S_{2}[/tex] = u t + [tex]\frac{1}{2}[/tex] × [tex]a_{2}[/tex] × [tex]t^{2}[/tex]

⇒ [tex]S_{2}[/tex] = 10 × 2 - [tex]\frac{1}{2}[/tex] × 5 × [tex]2^{2}[/tex]

⇒ [tex]S_{2}[/tex] = 20 - 10

⇒ [tex]S_{2}[/tex] = 10 mi

This is the distance traveled by the car in the next two hours.

Thus the total distance traveled by the car = [tex]S_{1}[/tex] + [tex]S_{2}[/tex]

S = 5 + 10 = 15 mi

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