P5 Acceleration (page 208) Flashcards
What is Uniform acceleration?
its just speeding up (or slowing down) at a constant rate.
What is Acceleration?
it is how quickly you’re speeding up
Is Acceleration the same as velocity or speed?
No
Explain what the Accerlation is?
it is the change in velocity in a certain amount of time.
What equation can you use to find the average acceration of an object is?
a = Δv
___
t
a - acceleration (m/s²)
Δv - Change in velocity (m/s)
t - Time (s)
see diagram 1 on page 208.
A cat accelerates at 2.5 m/s² from 2.0 m/s to 6.0 m/s.
Find the time it takes to do this?
t = Δv ÷ a
(6.0 - 2.0) ÷ 2.5 - 1.6 s
What do Deceleration mean?
its just negative acceleration (if something slows down, the change in velocity is negative).
You need to be able to estimate accelerations.
To do this you need the typical speeds from page 207.
A car is travelling along a road, when it collides with a tree and comes to a stop. Estimate the deceleration of the car?
1) first, give a sensible speed for the car to be travelling at.
2) next estimate how long it would take the car to stop
3) put these numbers into the acceleration equation
4) the question asked for the deceleration, so you can lose the minus sign (which shows the car is slowing down)
e.g.
the typical speed of the car is ~ 25 m/s
the car comes to a ctop in ~1 s.
a = Δv ÷ t
= (-25) ÷ 1
= -25 m/s²
so the deceleration is ~25 m/s²
(the ~ symbol just means it’s an approximate value (or answer).
What do ‘Uniform’ Acceleration mean?
a constant Acceleration
(a Constant Accelleration is sometimes called uniform acceleration.
Acceleration due to gravity (g) is uniform for objects in free fall.
Its roughly equal to 0.8 m/s² near the Earth’s surface and has the same values as gravitational field strength (p.202).
What is the equation for uniorm acceleration?
v² - u² = 2as
V² - final velocity (m/s)
a - Acceleration (m/s²)
s - Distance (m)
u - initial velocity (m/s)
(initial velocity is just the starting velocity of the object).
A van travelling at 23 m/s starts deceleratng uniformly at 2.0 m/s² it heads towards a built-up area 112 m away. What will its speed be when it reaches the build-up area?
1) First, rearrange the equation so v² is on one side
2) Now put the numbers in - remember a is negative because it’s a deceleration
3) finally, square root the whole thing.
V² = u² + 2as
V² = 23² + (2 × -2.0 x 112)
= 81
V = √81 = 9 ms
You might not be told what equation to use in the exam, so make sure you can spot when to use the equation for uniform acceleration. Make a list of the information you’re given to help you see what to do.
A ball is dropped from a height, h, above the ground. The speed of the ball just before it hits the ground is 7 m/s. Calculate the height the ball is dropped from. (aceleration due to gravity ≈ 9.8 m/s² (3 marks)
u = 0 m/s, v = m/s, a = g = 9.8 m/s²
s = (v² - u²) + 2a (1 mark)
= (49 - 0) + (2 × 9.8) (1 mark)
= 2.5 m (1 mark)
