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Physics_is_beautiful

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- 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.

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.

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