Answer:
B.
Explanation:
A position-time graph is a graph of the position of an object against (versus) time. The slope of the line of a position-time graph is typically used to determine or calculate the velocity of an object.
Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.
Mathematically, velocity is given by the equation;
Uniform velocity can be defined as a situation in which a physical object or body travels equal distances in equal intervals of time along a fixed direction, regardless of how small the interval may be.
During uniform velocity, both the magnitude and direction of the physical object in motion do not change with respect to time. Also, an object in uniform velocity has zero tangential acceleration and its graph is a straight line parallel to the time axis.
Hence, a vehicle with a uniform velocity of 10 meters per seconds (m/s) is represented by Graph B.
Answer:
1.5 km/s²
Explanation:
Given that:
a car starts from rest; it means the initial velocity (u) = 0 km/hr = 0 m/s
after time (t) = 20 seconds
the final velocity = 108 km/hr = 30 m/s
The acceleration (a) of the car can be determined by using the formula:
a = 1.5 km/s²
Answer:
Explanation:
This is an energy conservation problem in that the total energy available to a system s constant trhoughout the whole problem. That is, in equation form:
TE = PE + KE. This is a bit of a problem due to sig figs here, but I will keep with the rules for them all the way up til the end of the problem (I'll tell you when I veer away from the rule and why I did when I get there).
We first need to find the total energy available to the system and then use that value throughout the rest of the problem.
TE = PE + KE where
PE = mgh (mass times gravity times height of the object) and
KE = (one-half times the mass of the object times the square of the velocity of the object). We solve for PE first, rounding to the correct number of sig figs:
PE = 97(9.8)(3.1) and
PE = 2900 J (2 sig figs here since all the numbers given have 2 sig figs in them). Now for KE
which, to 2 sig figs (notice I added a .0 to the 8), is
KE = 2200. This means that the total energy available to the sled throughout the whole trip is
TE = 2900 + 2200 so
TE = 5100 J
Now for the second part of the problem, the sled is at a different height so the PE is different, and we are asked to find the velocity at the top of this second hill. The total energy equation then is
5100 = PE + KE. Solving for PE first:
PE = (97)(9.8)(1.2) and, to 2 sig figs:
PE = 82. Now for the KE:
. Plugging all of that into the total energy equation gives us:
and isolating the v:
and this is where I veer away from the sig figs. Doing the subtracting first, as we should, means that we would round to the hundreds place. Since the least significant place in the number 82 is the tens place and the least significant place in the 5100 is in the hundreds place, we round to the hundreds place. But I didn't keep to this rule because I'm not sure how adamant your physics teacher is about sig figs. I'll give you the answer that I would expect from my students (the right one) after I give you this one. I just simply subtracted 5100 - 82 and used the answer the calculator gave me which was 5018:
gives us that the velocity at the top of the 1.2 meter hill is
10.2 m/s.
Doing this the right way, using the rules for sig figs properly:
gives us, to 2 sig figs,
v = 1.0 × 10¹ m/s