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 did you get t=81/3

Good

Well, well, well, looks like we have a case of a decelerating car! Time to put my clown hat on and tackle these questions. Hold on tight!

So, we have an initial velocity of 25 m/s and a constant deceleration of 3 m/s². Let's start with the first question: the velocity of the car when t = 4s.

Using the equation of motion v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can plug in the values we know:

v = 25 m/s - 3 m/s² * 4 s

Now, let me calculate that for you... drumroll please...

[Calculating...]

And the answer is... 13 m/s! So, when t = 4s, the velocity of the car is 13 m/s. It's like going from a speeding bullet to a slow snail!

Now, for the second question: the displacement of the car during the 4-s time interval.

To find the displacement, we can use the equation of motion s = ut + (1/2)at², where s is the displacement. We already know the initial velocity (u) and the time (t), but we need the acceleration (a). In this case, since the car is decelerating, the acceleration would be negative.

Plugging in the values, we have:

s = 25 m/s * 4 s + (1/2)(-3 m/s²)(4 s)²

Time for another thrilling calculation...

[Calculating...]

And the answer is... -32 m! The displacement of the car during the 4-s time interval is -32 m. So, clown's honor, the car moves backwards by 32 meters. Maybe the driver hit his clown nose on the brake pedal too hard!

Lastly, for the third question: how much time is needed to stop the car?

To stop the car, it would mean that the final velocity (v) is 0 m/s. We can use the same equation v = u + at here, but this time we need to solve for time (t).

So, let's set up the equation:

0 m/s = 25 m/s + (-3 m/s²) * t

Now, let me do some magical calculations...

[Calculating...]

And, ta-daaa! The answer is... 8.33 seconds! It would take approximately 8.33 seconds to stop the car. That's like waiting for the punchline of a never-ending clown joke!

So there you have it: the velocity when t=4s is 13 m/s, the displacement during the 4-s time interval is -32 m, and the time needed to stop the car is around 8.33 seconds. Hope that brought a smile to your face!

To determine the velocity of the car after 4 seconds, we can use the equation of motion:

v = u + at

Where:
v = final velocity (unknown)
u = initial velocity = 25 m/s
a = acceleration (deceleration in this case) = -3 m/s^2 (negative because it's decelerating)
t = time = 4 seconds

Plugging in the values, we get:

v = 25 m/s + (-3 m/s^2) * 4 s
v = 25 m/s - 12 m/s
v = 13 m/s

Therefore, the velocity of the car after 4 seconds is 13 m/s.

To calculate the displacement of the car during the 4-second time interval, we use the equation:

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

Where:
s = displacement (unknown)
u = initial velocity = 25 m/s
t = time = 4 seconds
a = acceleration (deceleration in this case) = -3 m/s^2

Plugging in the values, we get:

s = (25 m/s)(4 s) + (1/2)(-3 m/s^2)(4 s)^2
s = 100 m - (1/2)(-3 m/s^2)(16 s^2)
s = 100 m - (1/2)(-48 m)
s = 100 m + 24 m
s = 124 m

Therefore, the displacement of the car during the 4-second time interval is 124 meters.

To find out how much time is needed to stop the car, we can use the equation:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s, since the car needs to stop)
u = initial velocity = 25 m/s
a = acceleration (deceleration in this case) = -3 m/s^2
s = displacement (unknown)

Plugging in the values, we get:

(0 m/s)^2 = (25 m/s)^2 + 2(-3 m/s^2)s
0 m/s = 625 m^2/s^2 - 6 m/s^2 s
6 m/s^2 s = 625 m^2/s^2
s = 625 m^2/s^2 / 6 m/s^2
s = 104.1667 m

Therefore, the car needs a displacement of approximately 104.1667 meters to stop completely.

However, since the displacement during the 4-second time interval was already determined to be 124 meters, it means that the car would have overshot the stopping point within this time frame. So, to calculate the exact time needed to stop the car, further information or calculations are required.

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