3.1 - 3.8

Question 1

whats the gradient of a distance-time graph equal to

Answer 1

the speed

Question 2

how is average speed found

Answer 2

the total distance travelled / total time taken

Question 3

how is instantaneous speed found

Answer 3

drawing a tangent on a distance-time graph at the time you want to know its speed at then finding its gradient

Question 4

define instantaneous speed

Answer 4

the speed at the moment it is measured

Question 5

define average speed

Answer 5

the rate of change of distance over an entire journey

Question 6

define scalar

Answer 6

a quantity with only magnitude

Question 7

define vector

Answer 7

a quantity with magnitude and direction

Question 8

how can average velocity be found

Answer 8

the displacement/ total time taken

Question 9

define velocity

Answer 9

the rate of change of displacement

Question 10

define acceleration

Answer 10

the rate of change of velocity

Question 11

what can be determined from a velocity-time graph

Answer 11

acceleration: the gradient
displacement: the area under the curve

Question 12

how can the area of a non-linear graph be determined

Answer 12

by counting the squares that are completely or mostly beneath the curve

Question 13

what are the SUVAT equations

Answer 13

v= u + at
v^2 = U^2 + 2as
s= ut +1/2 at^2
s= vt -1/2 at^2
s=1/2(u+v)t

Question 14

how can the v=u+at equation be derived

Answer 14

from a velocity-time graph where the initial velocity is u and the final velocity is v. the gradient = (v-u)/t. the gradient is the acceleration so this can be rearranged to get v=u+at

Question 15

how can the s=ut+1/2(at^2) equation be derived

Answer 15

from a velocity-time graph with initial velocity u and final velocity v. split it into a triangle and rectangle. the area of the rectangle is ut and the triangle is 1/2(v-u)t.
from v=u+at you get v-u=at. sub that in to get 1/2at^2. add them together for the total area which equals the displacement(s). s=ut + 1/2(at^2)

Question 16

how can the s=1/2(u+v)t equation be derived

Answer 16

from a velocity-time graph with initial velocity u and final velocity v. take the area of the trapezium to get 1/2(u+v)t. the area is the displacement(s) so s=1/2(u+v)t

Question 17

how can the v^2 = u^2 +2as equation be derived

Answer 17

rearrange v=u+at to get t=(v-u)/a. substitute that in to the equation s=1/2(u+v)t and rearrange to get v^2=U^2+2as.

Question 18

what is stopping distance

Answer 18

the distance travelled in the time it takes the driver to react to a hazard and the car to stop. the sum of the thinking distance and braking distance

Question 19

define thinking distance

Answer 19

the distance travelled between when a driver first perceives the need to stop and when the brakes are applied

Question 20

define braking distance

Answer 20

the distance travelled from the time the brakes are applied to the time the vehicle stops

Question 21

factors affecting stopping distance

Answer 21

vehicle speed,
brake conditions,
tyre conditions,
road conditions,
weather conditions,
driver alertness

Question 22

how does projectile motion work

Answer 22

horizontal and verticle components are independent. the horizontal speed remains the same because no force is acting on it. the vertical speed accelerates towards the ground due to gravity.

Question 23

define projectile motion

Answer 23

the motion of an object thrown into the air subject only to the force of gravity

Question 24

whats the shape of a displacement-time^2 graph for an object in free fall and why

Answer 24

a straight line through the origin. constant gradient. using s=ut +1/2at^2. u=0 and a=9.81 so it’s s=4.905t^2. 4.905 is constant so s is directly proportional to t^2