# GCSE stopping distances questionWatch

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Thread starter 2 years ago
#1
I have a question in a text book which asks me to calculate the stopping distance of a car from 30km/h.
I am told that the cars brakes can apply a force of 6.5 kN and that the car has a mass of 1000 Kgs.

However, I cannot get the answer given in the book at all.

This is how I work it out......

30 km/h = speed of 8.3 m/s

Acceleration (or maximum deceleration in this case) = Force / Mass Therefore 6500 Newtons / 1000 Kgs = 6.5 m/s squared

Using formula of Acceleration = Change in velocity / Time I rearrange the forumla for Time: Time = 8.3 / 6.5 = 1.28 seconds

Using the formula Distance = Speed X Time I get: 8.3 x 1.28 = 10.6 m to stop

But the book says 5m only?! And it doesn't explain how it gets 5m as the answer. 0
2 years ago
#2
Personally I would approach this from a energy conservation perspective. Calculate the energy the car has during its motion, you probably know a formula to calculate this, then use the fact the the work done in slowing the car down is force * distance. Using this approach you will probably find that the distance is more like 5m.

The problem with your method is that in your final step of distance = speed * time, you can't use the speed as 8.3m/s because during the deceleration the speed is constantly changing, and this formula only applied to constant speed.

Hope this Helps
0
2 years ago
#3
(Original post by Darwinion)
I have a question in a text book which asks me to calculate the stopping distance of a car from 30km/h.
I am told that the cars brakes can apply a force of 6.5 kN and that the car has a mass of 1000 Kgs.

However, I cannot get the answer given in the book at all.

This is how I work it out......

30 km/h = speed of 8.3 m/s

Acceleration (or maximum deceleration in this case) = Force / Mass Therefore 6500 Newtons / 1000 Kgs = 6.5 m/s squared

Using formula of Acceleration = Change in velocity / Time I rearrange the forumla for Time: Time = 8.3 / 6.5 = 1.28 seconds

Using the formula Distance = Speed X Time I get: 8.3 x 1.28 = 10.6 m to stop

But the book says 5m only?! And it doesn't explain how it gets 5m as the answer. Alternatively, if you are not familiar with energy conservation you could use the so-called SUVAT equations of motion. For example you have worked out the deceleration so you could use the equation v^2 = u^2 + 2*a*s where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance.

Hope this helps
0
Thread starter 2 years ago
#4
OK thanks for that!

Using your equation of v² = u² + 2as I actually do get a final value of 5.3 m

0 = 8.3² + (2 x 6.5 x s) Therefore 0 = 68.89 +13s Therefore rearranging for s = 68.89 / 13 = 5.3 m

But this damn book has not used that formula at all in the texts so far. Am I expected to be a mind reader? LOL In fact it's not listed in my books as a formula to learn.

I've just started self teaching for GCSE and only formula used so far are simple Distance = Speed x Time and Force = Mass x Acceleration
0
2 years ago
#5
Good job working it out. Be careful in your calculations though as the acceleration should be a negative number in this case (because deceleration acts in the opposite direction to the initial velocity).

Which textbook are you using? If the book doesn't mention these equations maybe the approach it wants to you take is similar to what you did initially but with a slight change:

"30 km/h = speed of 8.3 m/sAcceleration (or maximum deceleration in this case) = Force / Mass Therefore 6500 Newtons / 1000 Kgs = 6.5 m/s squaredUsing formula of Acceleration = Change in velocity / Time I rearrange the forumla for Time: Time = 8.3 / 6.5 = 1.28 seconds"

Then we can say Average speed = distance/time, so distance = average speed * time

The average speed when decelerating from 8.3m/s to 0m/s is (8.3 + 0)/2 = 4.15m/s

thus distance = 4.15m/s * 1.28s = 5.3m (which is approximately 5m, and the same as what we got above).

Hope this helps
0
Thread starter 2 years ago
#6
I have two books here: Collins Edexel IGCSE Physics and a CGP GCSE Physics workbook.

The workbook has a full page of all the formula used in the course and that equation isn't in there.
0
2 years ago
#7
I'd advise at least learning the energy method as in some situations it is much quicker to do. You'd say that the kinetic energy from the car (start) is converted into friction until it stops (end), using equations for these types of energy you get 1/2mv^2 = Fd, rearrange for d in this case.
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