Question is: a car travelling at 140 km/h decelerates for 20 secs. During this time it travels 500 m. What is its deceleration? What is its speed after deceleration?

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As I understand, the deceleration is uniform (the same all the time). Denote it as `agt0` and denote the initial speed as `V_0.`  In m/s `V_0 = 140/3.6.`

Then the speed is `V(t) = V_0 - a*t`  (note the minus sign), and the displacement is `D(t) = V_0 t -...

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Hello!

As I understand, the deceleration is uniform (the same all the time). Denote it as `agt0` and denote the initial speed as `V_0.`  In m/s `V_0 = 140/3.6.`

Then the speed is `V(t) = V_0 - a*t`  (note the minus sign), and the displacement is `D(t) = V_0 t - (a t^2)/2.`

It is given that `D(20) = 500,` therefore `V_0 * 20 - (a * 20^2)/2 = 500.` From this we find  `a = 2*(140/3.6*20-500) / 20^2 approx1.39 (m/s^2).`

Then we can find the speed after deceleration: it is

`V(20) = V_0 - a*20 approx 140/3.6 - 20*1.39 approx 11.1 (m/s).`

The value is positive which means the direction of movement remains the same.

The answers: the deceleration is about `1.39 m/s^2,` and the final speed is about `11.1 m/s.`

 

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