Accelerating and Braking a car

Classical Mechanics Level 2

Curves at high speed can be very dangerous, even if you're in a sports car with different systems such as stability control, traction control and abs brakes.

John accelerated his sportive car to a speed of 216 km/h, but the road is too sinuous and he had to reduce the speed of his car to 72 km/h in 3 seconds, approximately how much distance was traveled by John in this time?

120 m 147 m 150 m 180 m

Marcos Paulo Martins by Marcos Paulo Martins , 23, Brazil

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3 solutions

Raiyun Razeen
May 29, 2014

216 km/h = 216000/3600 ms^-2 = 60 ms^-2

72 km/h = 72000/3600 ms^-2 = 20 ms^-2

Now, u=60 ms^-2,v=20 ms^-2,t=3 s

So, a=(v-u)/t

or, a=(20-60)/3 = -13.33 ms^-2

We know, s= ut+(1/2)at^2

So, s= 60 3+0.5 (-13.33)*3^2

or, s= 180 - 59.985 = 120.015

Therefore the answer is 120

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Raj Magesh
Jun 1, 2014

Assuming that the car decelerates uniformly,

D i s t a n c e = t × v a v g Distance = t \times v_{avg}

216 k m h − 1 = 60 m s − 1 216 kmh^{-1} = 60 ms^{-1}

72 k m h − 1 = 20 m s − 1 72kmh^{-1} = 20 ms^{-1}

S = t ( u + v ) 2 S = \dfrac{t(u+v)}{2}

S = 3 s ( 20 m s − 1 + 60 m s − 1 ) 2 = 120 m S = \dfrac{3s(20ms^{-1} + 60ms^{-1})}{2} = \boxed{120m}

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Krishna Deb
May 28, 2014

used v=u+at n then s=ut+0.5gt^2

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