Revision sem1 q3

A truck and a car are moving with the same K.E. on a straight line road. If their engines are turned off at the same time, which one of them will stop at a lesser distance?

9 Answers

337
Sayan Sinha ·

Truck.

206
Sayantan Hazra ·

Can you explain why?....

337
Sayan Sinha ·

Same KE.

KE=0.5m2∞v2

So, if KE is constant for both the vehicles,
constant=0.5mv2
let constant KE be k
So,
→k=0.5mv2
k0.5=mv2
k0.51m=v2
→v21m
→m∞1v2

We assume mass of truck>car
So, more the mass lesser the velocity.
So, truck has lesser velocity.
So, truck will stop faster.
Also, weight of truck is more. Assuming we have same quality tyres for both the vehicles (i.e. coefficient of kinetic friction is same for both), friction=kN will be more for the truck. So more friction. This again adds to the retardation. So, truck stops faster.

Thanx...

206
Sayantan Hazra ·

since you have to compute which one stops at a lesser distance, compare the distances moved by them after engine is switched off......
Hint: Ffric = uN
→ ma = uN.........(if a be the retardation)

337
Sayan Sinha ·

Then the retardations are being same as...

ma=uN

N=mg (most of the time)

ma=umg
a=ug

If u is same for both, both will have equal retardation.

I hope my first part was correct where I showed m is inversely proportional to v2.
And, v2=u2-2aS
Here, v=0
For truck u is lesser.
u2=2aS
→lesser the velocity, lesser the distance as retardation is the same...

206
Sayantan Hazra ·

Very Good !! :)....another green for you!!

101
Sukrit Roy Chowdhury ·

if the truck and the car is opposed by same backward force which will stop first?

206
Sayantan Hazra ·

Calculate it yourself......and then see.. :)

Hint: Same breaking force, or backward force means
Ftruck = Fcar
or, m1a1 = m2a2

1
Asmita Chatterjee ·

Can we not say that assuming that the truck has more mass than the car, thus it will possess more inertia and thus it will stop at a farther distance than the car???.....if not then please tell why and please explain me the sum since i couldnt follow it...

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