Plzz answer this questions
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1. 19.28 secs

2. 154.22 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 16 m/s

Final velocity (v) = 0

Force (F) = 1000 N

Mass (m) = 1200 Kg

Time (t) =..?

Distance (s) =...?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Force (F) = 1000 N

Mass (m) = 1200 Kg

Acceleration (a) =.?

Force (F) = mass (m) x acceleration (a)

F = ma

1000 = 1200 x a

Divide both side by 1200

a = 1000/1200

a = 0.83 m/s²

Since the car is coming to rest, it means it is decelerating. Therefore, the acceleration is – 0.83 m/s²

1. Determination of time taken for the car to halt i.e stop. This can be obtained as follow:

Initial velocity (u) = 16 m/s

Final velocity (v) = 0

acceleration (a) = – 0.83 m/s²

Time (t) =.?

v = u + at

0 = 16 + (–0.83 x t)

0 = 16 – 0.83t

Rearrange

0.83t = 16

Divide both side by 0.83

t = 16/0.83

t = 19.28 secs.

Therefore, the time taken for the car to halt is 19.28 secs.

2. Determination of the distance travelled by the car before coming to rest. This can be obtained as follow:

Initial velocity (u) = 16 m/s

Final velocity (v) = 0

acceleration (a) = – 0.83 m/s²

Distance (s) =..?

v² = u² + 2as

0 = 16² + (2 x –0.83 x s)

0 = 256 – 1.66s

Rearrange

1.66s = 256

Divide both side by 1.66

s = 256/1.66

s = 154.22 m

Therefore, the distance travelled by the car before coming to rest is 154.22 m.

be subject to a force andchange its motion