Physics Speed Problem?

Question

I'm not looking for the answer to this so much as looking for what I need to do to get the answer. So please explain to me how I go about it. Thanks!

Part 1

A particle has initial horizontal Velocity of 2.9 m/s and initial upwards velocity of 4.9 m/s. It then has a horizontal acceleration of 1.5 m/s^2 and downward acceleration of 2.1m/s^2. What is the speed after 1.8 s? Answer in m/s.

Part 2

What is the direction of its velocity at this time with respect to the horizontal? Answer between -180 and +180 degrees

Answer 1

Each component can be looked at separately using
vx(t)=vx0+ax*t

vy(t)=vy0+ay*t

The resultantant vector has magnitude of
sqrt(vx^2+vy^2)
and direction
ATAN(vy/vx)

be careful with signs

vx(t)=2.9+1.5*t
=5.6 m/s
vy(t)=4.9-2.1*t
=1.12 m/s

magnitude
5.71 m/s
angle
11.31 degrees

j

Answer 2

Hi,
Well, frankly, it'd be easier to work the problem than to try to tell you enough without giving too much away. :-) But here goes:

1) In the last step you need to know the horizontal and vertical components of the velocity at time = 1.8 s so you can calculate the speed from those two components. If you're familiar with vectors, the speed will be the vector sum of the two components. If you don't know vectors, you can just think of the velocities as two legs of a right triangle.
2) Now, to back up a little, you need to know that the both the horizontal and vertical velocities have both a constant component, the initial velocity, and a time dependent component, acceleration multiplied by time. You need to know whether the time component and constant component add or whether one is subtracted from the other. Don't try to make this part too complicated. The equations are quite simple.
3) Just a caution: There's somewhat of an ambiguity in the problem as stated, in that particles close to the earth have a downward acceleration of -9.8 m/s². I suppose that we could assume that your particle is out in space somewhere, but read your problem carefully and consider it's context to see if gravity is to be considered.

For Part B, the answer should be obvious from the sign of the vertical component of the velocity.

Hope this is of some help.
FE

Answer 3

Part 1

Your particle has two velocity components that work independantly because they are perpendicular to each other. So, break down the question into two parts: Vertical and Horizontal.

So your starting vertical velocity is +4.9 m/s (positive because it is headed upwards)

Horizontal velocity is +2.9 m/s (we're assuming it's headed in the positive direction to make the math easier)

Remember that V(final) = V(initial) + a*t

The accleration for the vertical is downward, so it would be
-2.1 m/s^2

I'll leave the math to you :)


Part 2

Your velocity has two components an x and a y

You are looking for an angle with respect to the horizontal. Angles are related to the x and y components of a velocity, or any quantity for that matter, by the equation:
angle = arctan(y/x)

Remember the bounds of the arctan function and make sure the angle you get fits in the quadrant that your y and x values would put it on a coordinate plane. For example, if both are positive, you should expect an angle between 0 and 90 degrees.

Answer 4

Good for you, not expecting someone else to do all the work!

Treat each dimension independently. There are two separate cases in which there is an initial velocity, then an acceleration that acts for a given time. Just be sure to pay attention to whether the acceleration increases or decreases the velocity.

Once you have the vertical and horizontal velocities, add them as vectors to get the magnitude (speed) and direction (angle) of the resulting vector.

Answer 5

1. Ur speed is 110 kmph is equal to 30.55 m/s. So u trawell 30.55 m in a second . Ur answer in 2 seconds is double of it. It's nearly 61m 2. (a) U r going with 65 mph and drove 130 miles so the time taken to drive 130 miles is 2 hours coz u trawell 65 miles in an hour. Now subtract these two hours from total time. Remaining time is 1hour 20 min. Which is 1.33 hours. During these 1.33 hours ur speed was 55mph. so trawelled distance is equal to (speed x time ) so u trawelld 73.33 miles. Total distance is 130 miles + 73.33 miles answer is 203.33 miles (b) average velocity is total distance trawelled devided by the total time. Avg. Velocity = 203.33/3.33 ~ 61 mph