Always follow the traffic rules!

Classical Mechanics Level 2

At a distance of $500\text{ m}$ from the traffic light, brakes are applied to an automobile moving with a velocity of $20\text{ ms}^{-1}$ .

Find the position (in meters) of the automobile with respect to traffic light 50 seconds after applying the brakes, if its acceleration is $-0.5\text{ ms}^{-2}$ before it comes to rest.

100 125 375 400

1 solution

Sam Bealing
Jun 20, 2016

We can use $suvat$ equations. As the car is braking we can assume its velocity does not drop below $0$ .

$(0.5) \times 50=25 >20$

So we can assume the car reaches a complete stop and hence $v=0$ .

$s=?,v=0,u=20,a=-0.5$

$v^2=u^2+2as \implies 0^2=20^2+2(-0.5)s \implies s=400 \; m$

So the total distance travelled is $400 \; m$ making the car $500-400=\boxed{\boxed{100 \; m}}$ away from the traffic light.

Moderator note:

The svuat equations only hold when acceleration is constant. When the car has stopped, there is no further deceleration, so the equations will not hold over the entire time period. We have to find the section of time when the conditions are satisfied, in order for the equation to hold.

You cannot assume the final velocity after 50s is 0 , in-deed there is nothing mentioned that that car stops after 50s.

$\text{s = ut + }\frac{1}{2}\text{(at}^{2}\text{)} \\ => s = 20 * 50 + \frac{-0.5 * 50 * 50}{2} \\ => s = 375m$

So, if you need to use 3rd equation of motion, you need to find v .

$v = u + at = 20 + -0.5 * 50 = -2ms^{-1}$

[CLARIFIED : ANSWER IS CORRECT]

Viki Zeta - 4 years, 12 months ago

I think however we can assume that when you brake, your car doesn't doesn't start accelerating in the opposite direction after you stop. I have clarified my solution. I do think the question could maybe do with rewording though.

Sam Bealing - 4 years, 12 months ago

I think Vicky Vignesh has got a point in his arguement. The question does not state the car comes to rest after 50s. if the car would have come to rest after 50s then, there would have been no point in giving the time interval. I think the correct ans is 125m.

Arjun Ramesh - 4 years, 12 months ago

@Arjun Ramesh – I do think he has a point, it just seemed illogical to me that you could apply the brakes then start reversing. I think we probably need Rishabh Tiwari to clarify his question.

Sam Bealing - 4 years, 12 months ago

@Sam Bealing – This question is not that hard , for which Vicky Vignesh is thinking too deep. This is basically about the fact that if you are able to notice that the car comes to rest 10 s before the given time interval of 50 s.

Moreover , thinking practically , (I agree with Sam) , its very odd to think that the car starts reversing after applying brakes.

Its not about a particle or a helicopter , that you can use the $2^{nd}$ $eq^{n}.$ of motion for displacement . Its just a practical life observation.

Hence the answer is $\color{#3D99F6}{\boxed {100 m}}$ & Sam is absolutely correct.

Hope it helps . Thank you :-)

Rishabh Tiwari - 4 years, 12 months ago

@Rishabh Tiwari – It is illogical to give time interval in this question , please remove it .

Sabhrant Sachan - 4 years, 12 months ago

@Sabhrant Sachan – I think it matters so that the reader gets confused! For fun I think it should be there.:-)

Rishabh Tiwari - 4 years, 12 months ago

Perfect solution , +1!

Rishabh Tiwari - 4 years, 12 months ago

Good solution. Yeah it's highly impractical for a car to go in reverse with help of breaks. LOL.

You could have mentioned that, using 1st eqn of motion, assuming car has stopped and v = 0. So the time in seconds is 40s.

Viki Zeta - 4 years, 12 months ago

but if i apply s= ut+1/2ft^2 then how come the result will be 100 because S= 375 which means the car is away from the traffic light would be 500 - 375 = 125

Parikshit Das - 4 years, 11 months ago

Always follow the traffic rules!

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