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If the equation of an average velocity of a sport car was given by the equation V(t) = 3t^2 -6t +24. Determine its displacement

alisha [4.7K]

Answer:

72

Explanation:

The displacement of an object can be found from the velocity of the object by integrating the expression for the velocity.

In this problem, the velocity of the sport car is given by the expression

$v(t)=3t^2-6t+24$

In order to find the expression for the position of the car, we integrate this expression. We find:

$x(t)=\int v(t) dt=t^3-3t^2+24t+C$

where C is an arbitrary constant.

Here we want to find the displacement after 3 seconds. The position at t = 0 is

$x(0)=0^3-0+0+C=C$

While the position after t = 3 s is

$x(3)=3^3-3(3)^2+24(3)+C=72+C$

Therefore, the displacement of the car in 3 seconds is

$d=x(3)-x(0)=72+C-C=72$

7 0

3 years ago

I got part c right but idk why the other parts are wrong HELP!

dedylja [7]

a) The impulse is 76.5 Ns

b) The average force is 546.4 N

c) The final speed is 31.5 m/s

Explanation:

a)

The impulse exerted on an object is defined as

$J=\int F\Delta t$

where

F is the magnitude of the force exerted on the object

$\Delta t$ is the time interval during which the force is applied

If we consider a graph of the force applied vs time, it follows that the impulse exerted is equal to the area under the graph.

Therefore, in this problem, we can calculate the impulse by computing the area under the graph. We have a trapezium, whose bases are

$B=0.14-0 = 0.14s\\b=8-5=3s$

and whose height is

$h=900 N$

Therefore, the area (and the impulse) is

$J=\frac{(B+b)h}{2}=\frac{(0.14+0.03)(900)}{2}=76.5 Ns$

b)

In this problem, the force applied is not constant. However, we can rewrite the impulse also as

$J=F_{avg} \Delta t$

where

$F_{avg}$ is the average force exerted during the whole time $\Delta t$

In this problem we have

J = 76.5 Ns is the impulse (calculated in part a)

$\Delta t = 0.14 s$ is the time interval

Solving for the average force, we find

$\Delta t = \frac{J}{F_{avg}}=\frac{76.5}{0.14}=546.4 N$

c)

According to the impulse theorem, the impulse exerted on an object is equal to the change in momentum of the object:

$J=\Delta p = m(v-u)$

where

m is the mass of the object

v is the final velocity

u is the initial velocity

In this problem, we have

J = 76.5 Ns

m = 3.0 kg is the mass

u = 6.0 m/s is the initial velocity

Solving for v, we find the final velocity (and speed):

$v=u+\frac{J}{m}=6.0+\frac{76.5}{3}=31.5 m/s$

Learn more about impulse and momentum:

brainly.com/question/9484203

#LearnwithBrainly

6 0

3 years ago

An student in a swimming pool weighs 450 N and in a regular balance out of the swimming pool weighs 700 N. What is the value of

rewona [7]

The value of buoyant force is 700-450=300N

7 0

3 years ago

If a ball rolls at constant velocity, it is rolling at constant speed? explain

Shtirlitz [24]

Velocity is speed with direction. So, if velocity varies directly with speed, that statement would be true. A constant velocity would resort in a constant speed. They are connected and are dependant on each other.

I hope this helps!
~kaikers

7 0

3 years ago

You have an initial velocity of -3.0 m/s. You then experience an acceleration of 2.5 m/s2 for 9.0s; what

Tatiana [17]

27.9 idkkkk look it up on photomath

8 0

3 years ago