A driver in a moving car applies the brakes. The car slows to a final speed of 2.60 m/s over a distance of 40.0 m and a time interval of 8.45 s. The acceleration while braking is approximately constant.
a) What is the car's original speed before braking?
(b) What is its acceleration during this time? (The car's initial velocity is in the positive direction. Indicate the direction with the sign of your answer.)

The rate of change of displacement is known as velocity. The rate at which velocity changes is called acceleration. Due to the fact that it includes both magnitude and direction, velocity is a vector quantity. As the rate at which velocity changes, acceleration is likewise a vector quantity.
The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is accelerating. As a vector, velocity has both direction and magnitude.

Given ;

over a distance of 40.0 m

[tex]\Delta x = 40 m[/tex]

a time interval of 8.45 s

[tex]\Delta t = 8.45 sec[/tex]

final speed of 2.60 m/s

V = 2.60 m/s

for constant acceleration

[tex]\Delta x= \frac{v + u }{2} \times t[/tex]

40 = 1/2 (2.6 + u ) * 8.45

2.6 + u = 80 /8.45 =9.47

u = 9.47 - 2.6 = 6.87 m/s.

the car's original speed before braking is 6.87 m/s

acceleration :

a =( v - u ) /t

a = (2.6 - 6.87 ) /8.45

a = -0.51 m/s².

acceleration during this time is -0.51 m/s².

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