-A car accelerates uniformly from rest to a speed of 22.3 km/h in 4.8 s.
Find the distance it travels during this time.
-A driver in a car traveling at a speed of 67 km/h sees a cat 117 m away on the road.
How long will it take for the car to acceler- ate uniformly to a stop in exactly 113 m?
Answer in units of s
Find the distance it travels during this time.
-A driver in a car traveling at a speed of 67 km/h sees a cat 117 m away on the road.
How long will it take for the car to acceler- ate uniformly to a stop in exactly 113 m?
Answer in units of s
-
First, you look at the kinematic equations.
Then you use the one that allows you to calculate what you want to find.
So, here's the main four:
(note: d is distance, u is the initial velocity, v is the final velocity, a is acceleration, and t is time.)
d = ut + (at^2)/2
v^2 = u^2 + 2ad
v = u + at
d = (u + v)t/2
In the first problem, you want to find distance. You have the initial velocity (0), the final velocity (22.3 km/h) and time (4.8 seconds)
--- note: I think you put down the wrong final velocity --- 22.3 m/sec sounds more feasible than 22.3 km/h. If you are right, then you need to convert the km/h into m/s
Anyway, this sounds like a use for the fourth equation (I'll assume you meant 22.3 m/s to illustrate)
d = (u + v)t/2
==> d = (0 + 22.3) * 4.8 /2
==> d = 22.3 * 2.4 = 53.52 metres
For the second problem, you have the initial velocity, the final velocity and the distance. and you want to find the time. Again, it's the fourth equation, but you need to change it to find the time.
d = (u + v) * t /2
==> 2d = (u + v) * t
==> t = 2d/(u + v)
Plug in the numbers, and do the calculations
(again: note that your initial velociity as you stated it is in km/h. Should this be m/s? If you have stated the problem right, you fisrt need to convert the km/h into m/s .... 1 km/h = 1000/3600 = 10/36 = 5/18 m/s....)
Then you use the one that allows you to calculate what you want to find.
So, here's the main four:
(note: d is distance, u is the initial velocity, v is the final velocity, a is acceleration, and t is time.)
d = ut + (at^2)/2
v^2 = u^2 + 2ad
v = u + at
d = (u + v)t/2
In the first problem, you want to find distance. You have the initial velocity (0), the final velocity (22.3 km/h) and time (4.8 seconds)
--- note: I think you put down the wrong final velocity --- 22.3 m/sec sounds more feasible than 22.3 km/h. If you are right, then you need to convert the km/h into m/s
Anyway, this sounds like a use for the fourth equation (I'll assume you meant 22.3 m/s to illustrate)
d = (u + v)t/2
==> d = (0 + 22.3) * 4.8 /2
==> d = 22.3 * 2.4 = 53.52 metres
For the second problem, you have the initial velocity, the final velocity and the distance. and you want to find the time. Again, it's the fourth equation, but you need to change it to find the time.
d = (u + v) * t /2
==> 2d = (u + v) * t
==> t = 2d/(u + v)
Plug in the numbers, and do the calculations
(again: note that your initial velociity as you stated it is in km/h. Should this be m/s? If you have stated the problem right, you fisrt need to convert the km/h into m/s .... 1 km/h = 1000/3600 = 10/36 = 5/18 m/s....)