Help with question on motion: Avoiding a rear-end collision

[Physics_is_beautiful]

Homework Statement

The question : A car and a truck are both traveling with a constant speed of 20 m/s. The car is 10 m behind the truck. The truck driver suddenly applies his brakes, causing the truck to slow to a stop at the constant rate of 2 m/s2. Two seconds later, the driver of the car applies their brakes and just manages to avoid a rear-end collision. Determine the constant rate at which the car slowed.

Relevant Equations

basic S,U,T,A,V equations

The question : A car and a truck are both traveling with a constant speed of 20 m/s. The car is 10 m behind the truck. The truck driver suddenly applies his brakes, causing the truck to slow to a stop at the constant rate of 2 m/s2. Two seconds later, the driver of the car applies their brakes and just manages to avoid a rear-end collision. Determine the constant rate at which the car slowed.

How I tried to solve it :
--> if truck decelerates at 2 m/s^2, then it covers s = ut+0.5at^2 distance by the time it stops.
= (20)(10)+(0.5)(-2)(100)
= 100m

--> however, the car is 10m behind the truck, thus the car travels 110m ( for the sake of calculation, we assume that 'nearly collide' = occupying the same space, as without the length of the truck, it's not feasible to calculate the distance for 'nearly colliding'.)

--> Therefore, we now know that the car decelerates at some speed, in some time, covering 110-40m = 70m (as the car starts decelerating 2 seconds after) before coming to a stop.

--> next, using avg speed = (v+u)/2 = 20/2 = 10m/s,
we calculate the time taken by :
time = 70m/10 m/s = 7s

--> Thus, we now calculate for a by a = (v-u)/t
= -20/7
= -2.8 ms^-2

However, when checking the answer script, the answer was given as -3.33 ms^-2
acceleration.

I then tried solving using v^2 = u^-2as, only to get - 2.8 approx. as the acceleration.

I then attempted to solve it by adding in the length of an average truck ( approx = 8.5m), substitute into formula, to get

0 = 400-123a
leading to a = -3.3

However, Had this been the case, they would have given the length of truck in the question, so I doubted this would be the correct solution.

 

[Chestermiller]

The car has 8 seconds to travel 70 meters, starting at 20 m/sec?

 

[Halc]

( for the sake of calculation, we assume that 'nearly collide' = occupying the same space, as without the length of the truck, it's not feasible to calculate the distance for 'nearly colliding'.)

Yes, but the distance between the vehicles is from rear of truck to front of car, so length of either vehicle is irrelevant.

--> Therefore, we now know that the car decelerates at some speed, in some time, covering 110-40m = 70m (as the car starts decelerating 2 seconds after) before coming to a stop.
--> next, using avg speed = (v+u)/2 = 20/2 = 10m/s,

Agree up to here, sort of. This is the calculation of how far the car will move before coming to a stop, but the collision doesn't occur at that point if the plan is to stop there. The car must brake hard enough to be zero distance from the truck when the two match speeds.

we calculate the time taken by :
time = 70m/10 m/s = 7s

This is assuming an answer to the question asked, and a wrong one at that.
How long does the car have to come to a stop once the brakes are applied?
Better question: How long does it have to match speed with the truck once the brakes are applied?
The answer is the shorter time of the two.

 

[haruspex]

covering 110-40m = 70m (as the car starts decelerating 2 seconds after) before coming to a stop.

You are making an unjustified assumption about the possible ways a collision might occur.

 

[Steve4Physics]

= -20/7
= -2.33 ms^-2

But 20/7 = 2.8 approx., not 2.33!

However, when checking the answer script, the answer was given as -3.33 ms^-2
acceleration.

I then tried solving using v^2 = u^-2as, only to get - 2.8 approx. as the acceleration.

So your 2 methods agree (as they should!). I think -2.8m/s² (approx.) is correct and the officiaL answer (-3.33m/s²) is wrong. But no one else has suggested this, so maybe I'm missing something.

Edit. The reason why -2.8m/s² is wrong is that it assumes that the car just avoids collision with the stationary truck. But you need to make sure that the car doesn't hit the truck while the truck is still decelerating. @haruspex hinted at this in Post #4

 

[haruspex]

maybe I'm missing something.

Avoiding the vehicle in front involves more than merely coming to rest at the same point.

 

[Steve4Physics]

Avoiding the vehicle in front involves more than merely coming to rest at the same point.

Yes - I'd already edited my post.