A car has an initial speed of 25m/s and a constant deceleration of 3m/s^2. Determine thewvelocity of the car when t=4s. What is the displacement of the car during the 4-s time interval? How much time is needed to stop the car?

How do i get the displacement?

I already did this.

distance = 19*4 = 76 meters

v = Vi + a t
v = 25 - 3 t
v = 25 - 3(4)
v = 13 m/s after 4 sec
average speed during the 4 s = (25+13)/2 = 19 m/s
so distance = 19*4 = 76 meters

0 = 25 - 3 t
t = 8 1/3 seconds

To calculate the displacement of the car during a 4-second time interval, you can use the equation of motion:

s = ut + (1/2)at^2

Where:
s is the displacement,
u is the initial velocity,
t is the time interval (in this case, 4s), and
a is the constant acceleration (or deceleration in this case).

Let's substitute the given values into the equation:

u = 25m/s (initial speed)
t = 4s (time interval)
a = -3m/s^2 (deceleration, negative sign indicates deceleration)

s = (25m/s)(4s) + (1/2)(-3m/s^2)(4s)^2
s = 100m - 12m
s = 88m

Therefore, the displacement of the car during the 4-second time interval is 88 meters.

Now, let's calculate the time needed to stop the car. When the car comes to a stop, its final velocity (v) will be 0 m/s. Using the equation of motion:

v = u + at

Where:
v is the final velocity,
u is the initial velocity,
a is the acceleration (deceleration).

0 = 25m/s + (-3m/s^2)t_stop

Solving for t_stop:

25m/s = 3m/s^2t_stop
t_stop = 25m/s / 3m/s^2
t_stop ≈ 8.33s

Therefore, it will take approximately 8.33 seconds to stop the car.