Answer:
If a ball rolls on the ground describing a trajectory in a straight line and we take measurements about its position at different instants of time.
DISTANCE (m) 0 12 24 36
Time (s) 4 25 46 67
a) What is your speed?
b) What is its distance after 8 s?
c) What is its displacement after 8
Explanation:
Given that, the distance a ball as a function of time in table format.
So, we need to find the equation that governs the ball distance time graph
Since it is a straight line graph we will use straight line equation to get equation of the graph
y = mx + c
Where
m is the slope
And c is the intercept on y axis at t = 0
In this case, the y axis is distance 'd' and the x axis is time 't'
So, the slope can be calculated using
m = ∆y / ∆x = ∆d / ∆t
We can chose any two successive point, let choose the easy one
P1 = (t1, d1) = (4, 0)
P2 = (t2, d2) = (25, 12)
So,
m = (d2-d1) / (t2 - t1)
m = (12-0) / ( 25-4)
m = 12 / 21
So,
y = mx + c
y = d and x = t
d = 12t / 21 + C
Let find c taking any point, let taken at d=0 t = 4
0 = 12 × 4 / 21 + C
0 = 48/21 + C
C = -48/21
Then, the distance time graph is
d = 12t/21 - 48/21
Multiply through by 21
21d = 12t - 48
Divide through by 3
7d = 4t - 16
A. The speed can be calculated by taking the differential of the distance
d = 4t/7 - 16/7
Differentiation of the distance
s=d' = 4/7 m/s
So the speed is 4/7 m/s
Also, the speed is the slope of the distance time graph and the slope is already gotten to be
m = 12 / 21
m = 4/7 m/s
B. Distance after 8 seconds
At t = 8
7d = 4t - 16
7d = 4×8 -16
7d = 32-16
7d = 16
d = 16/7
d = 2.29m
C. Displacement after 8 second
Displacement is different from distance, displacement is a vector quantity
So, we need to find the initial displacement at t = 0
Then,
7d = 4t - 16
7d = -16
d = -16/7
d1 = -2.29m
So, at t = 8s, the displacement is calculated to be d2 = 2.29m
So, displacement is given as
D = ∆d = d2-d1
D = 2.29 - (-2.29)
D = 2.29 + 2.29
D = 4.58 m
So, the displacement after 8s is 4.58m
Spanish
Dado eso, la distancia de una pelota en función del tiempo en formato de tabla.
Entonces, necesitamos encontrar la ecuación que gobierna el gráfico de tiempo de distancia de la pelota
Como es un gráfico de línea recta, usaremos una ecuación de línea recta para obtener la ecuación del gráfico.
y = mx + c
Dónde
m es la pendiente
Y c es la intersección en el eje y en t = 0
En este caso, el eje y es la distancia 'd' y el eje x es el tiempo 't'
Entonces, la pendiente se puede calcular usando
m = ∆y / ∆x = ∆d / ∆t
Podemos elegir dos puntos sucesivos, elija el fácil
P1 = (t1, d1) = (4, 0)
P2 = (t2, d2) = (25, 12)
Entonces,
m = (d2-d1) / (t2 - t1)
m = (12-0) / (25-4)
m = 12/21
Entonces,
y = mx + c
y = d, x = t
d = 12t / 21 + C
Deje encontrar que c toma cualquier punto, déjelo tomado en d = 0 t = 4
0 = 12 × 4/21 + C
0 = 48/21 + C
C = -48/21
Entonces, el gráfico de tiempo de distancia es
d = 12t / 21 - 48/21
Multiplica por 21
21d = 12t - 48
Dividir entre 3
7d = 4t - 16
A. La velocidad se puede calcular tomando el diferencial de la distancia.
d = 4t / 7 - 16/7
Diferenciación de la distancia.
s = d '= 4/7 m / s
Entonces la velocidad es de 4/7 m / s
Además, la velocidad es la pendiente del gráfico de tiempo de distancia y la pendiente ya se ha determinado
m = 12/21
m = 4/7 m / s
B. Distancia después de 8 segundos
En t = 8
7d = 4t - 16
7d = 4 × 8-16
7d = 32-16
7d = 16
d = 16/7
d = 2.29m
C. Desplazamiento después de 8 segundos
El desplazamiento es diferente de la distancia, el desplazamiento es una cantidad vectorial
Entonces, necesitamos encontrar el desplazamiento inicial en t = 0
Entonces,
7d = 4t - 16
7d = -16
d = -16/7
d1 = -2.29m
Entonces, en t = 8s, el desplazamiento se calcula como d2 = 2.29m
Entonces, el desplazamiento se da como
D = ∆d = d2-d1
D = 2.29 - (-2.29)
D = 2.29 + 2.29
D = 4.58 m
Entonces, el desplazamiento después de 8s es 4.58m
