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Why is the stopping distance of a truck much shorter than for a train going the same

Why is the stopping distance of a truck much shorter than for a train going the same speed?
This question was in my physics book
Should I assume that these 2 are of different masses and the train is heavier ? How am I supposed to approach this ?
Original post by JanaALEVEL
Should I assume that these 2 are of different masses and the train is heavier ?

Yes
With old style trains that have the engine, motors and drive unit at one end, a train engine alone can be up to 100 tonnes, sometimes more, then you have to add the weight of every carriage it's pulling. With modern trains, the motors and drive units are much smaller but are under every carriage, so there's lots of them which all add up to lots of weight.
Trucks in the UK are typically under 50 tonnes fully loaded, except the specialist heavy haulage trucks.
Original post by Appirition
Yes
With old style trains that have the engine, motors and drive unit at one end, a train engine alone can be up to 100 tonnes, sometimes more, then you have to add the weight of every carriage it's pulling. With modern trains, the motors and drive units are much smaller but are under every carriage, so there's lots of them which all add up to lots of weight.
Trucks in the UK are typically under 50 tonnes fully loaded, except the specialist heavy haulage trucks.

I don't agree with this - greater masses are compensated by greater friction between wheels and the road or rails.

IMO sensible answers might be
few of the coaches on the train have brakes - this used to be the case especially on freight trains but I don't think it is any more... at least in the UK

the coefficient of friction between tarmac and rubber (truck tyres on road) is higher than the coefficient of friction between steel and steel (train wheels on rail)
Original post by JanaALEVEL
Why is the stopping distance of a truck much shorter than for a train going the same speed?
...

I think all they want is assume m (truck) << m (train), so for somewhat similar breaking force, deceleration of truck is much more than train's. Suvat: v^2 = u^2 + 2as => s = -u^2/2a = u^2/(2*decel). Same start speed u: Stopping Dist ∝ 1/decel => much shorter for truck.
(edited 4 years ago)

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