Some extremely strange physics questions! (1 Viewer)

[lookoutastroboy]

Hey guys,

I've got these 2 physics questions that I cannot do - if you know how to solve them, could you please post your full solution if possible =) Here goes -

1. A car 4.0 m long came up behind a semi-trailer 20.0m long travelling at a steady 72km/hr. After travelling behind the semi-trailer for some time, the driver of the car decided to overtake. If the car pulled out from behind the semi-trailer with the front of the car 10.0 m behind the rear of the trailer and accelerated at 3.5 ms^-2 until the rear of the car was 10.0 m in front of the truck, how far would the car travel and how long would it take?

2. A plane is flying parallel to the ground, with a velocity of 750 km h^-1, and at an altitude of 4.90 km. A lump of lead is dropped from the plane. How long before the lead reaches the ground (slightly easier, i think)



Thanks in advance and good luck,
lookoutastroboy

 

[scroates]

second one is easy.

using s=ut + at^2 but u=0

hence s=at^2

rearranging t=sqrt(s/a) where s=4900 metres and a= 9.8ms^-2

therefore answer is 22.36 seconds.

 

[addikaye03]

Hey guys,

I've got these 2 physics questions that I cannot do - if you know how to solve them, could you please post your full solution if possible =) Here goes -

1. A car 4.0 m long came up behind a semi-trailer 20.0m long travelling at a steady 72km/hr. After travelling behind the semi-trailer for some time, the driver of the car decided to overtake. If the car pulled out from behind the semi-trailer with the front of the car 10.0 m behind the rear of the trailer and accelerated at 3.5 ms^-2 until the rear of the car was 10.0 m in front of the truck, how far would the car travel and how long would it take?

2. A plane is flying parallel to the ground, with a velocity of 750 km h^-1, and at an altitude of 4.90 km. A lump of lead is dropped from the plane. How long before the lead reaches the ground (slightly easier, i think)



Thanks in advance and good luck,
lookoutastroboy

Q1)
Draw a diagram, showing dimensions. Use S=d/t.

Use the a-v relationship of a=(Vf-Vi)/t

Vf=Vi+at (Projectile motion formula)

I'll do it tomorrow if no '10ers have solved it.

 

[ar7]

Q1.
assuming both vehicles were in relative motion (72km/h or 20m/s)
i would add the distance needed to be travelled (10m(from behind the truck) + 20m(truck length) + 10m(front in front of truck) + 4m(rear is 10m in ahead of the truck) = 44m
a=d/t²
3.5= 44/t²
t²= 12.57
t= 3.55s

Q2.
∆y = Uyt + ½at²
4.9km = (0)(t) + 4.9 t²
4900 = 4.9t²
t² = 1000
t = 31.62s

horizontal component is not required