How Does an Incline Affect Truck Deceleration?

[scott86]

Not a homework question for me but one i was asked and would like to know how?

A truck traveling at 100km/h (27.7m/s) goes up a ramp inclined at 30', If it takes the truck 10 seconds to stop how far does the truck travel (m)?

Working out for a straight line is easy however i am unsure of the role the incline plays on this senario. Friction is not taken into account and the mass of the truck is not given

i don't neccessarily want answers just guidance please

Many thanks

 

[tiny-tim]

welcome to pf!

hiscott86! welcome to pf!

A truck traveling at 100km/h (27.7m/s) goes up a ramp inclined at 30', If it takes the truck 10 seconds to stop how far does the truck travel (m)?

there seems to be too much information here …

you can find the acceleration from the speed and the time, and it doesn't match the 30° slope

where does this question come from?​

 

[Doc Al]

A truck traveling at 100km/h (27.7m/s) goes up a ramp inclined at 30', If it takes the truck 10 seconds to stop how far does the truck travel (m)?

Seems like you're given too much information and that information is not consistent. (If you are not considering friction, then given the initial speed and angle of incline you should be able to figure out the distance and time.)

Is gravity the only decelerating force? If so, you can calculate the acceleration.

Another way is to calculate the average speed while the truck is slowing down.

 

[scott86]

The question is from a mates first year uni assignment we spent awhile on it and nothing we come up with seems logical, gravity is the only force to work with, the question was asking the horizontal distance of the ramp but you need the ramp distance in order to use trug to work out the horizontal, we got as far as using vectors but still unsure the rate of deceleration due to incline

 

[Gnosis]

Assumption: Earth gravity

If the values you provided are to pan out, then an additional force would be required.

For instance, if the truck were applying a given magnitude of torque to its wheels via its combustion engine as it ascends the 30 degree incline; then a 10 second ascent time is possible.

However, if the only force involved is Earth gravity, then a 10 second ascent time isn't possible, as the 30 degree incline per an initial velocity of 27.7 m/s would require just over 5.65 seconds to stop the truck’s ascent.

If this scenario were presented and it were stated that the truck does in fact stop its ascent in precisely 10 seconds, then the truck’s combustion engine is undoubtedly providing the additional required torque to extend the ascent time, as well as extend the ascent distance that it will traverse.

distance (s) traversed is:

s = v^2 / 2a

Since the acceleration (a) imparted by Earth gravity per a 30 degree incline per a free-wheeling vehicle is 4.9 m/s^2, the incline distance (s) traversed per an initial velocity (v) of 27.7 m/s is:

(27.7^2) / (2 * 4.9 m/s^2) = 78.29489796 meters

Per this derived distance (s) and the rate of acceleration (a) per the 30 degree incline, it’s realized per the following kinematics equation that a 10 second ascent time is not possible per gravity alone:

t = (2s / a)^.5

It therefore appears that the truck must be providing the necessary torque to extend distance traversed in order to accomplish the 10 second ascent time provided.