Solution :-
Given :
Distance 1 = 30 km
Distance 2 = 70 km
We know that speed = distance/time
and, Average speed = total distance/total time taken
When the train acquired a speed of 30 km/hr, the time taken = 30/30 = 1 hour
Average speed = 9distance 1 + distance 2)/(time 1 + time 2)
AS time 2 or t2 is time taken for the second part of the journey of 70 km
⇒ 40 = 100/(1 + t2)
⇒ 40 + 40t2 = 100
⇒ 40t2 = 100 - 40
⇒ 40t2 = 60
⇒ t2 = 60/40
⇒ t2 = 1.5
So, t2 or time taken to travel the second part of the journey is 1.5 hours.
Speed of the second part of the journey = distance 2/time 2
⇒ 70/1.5
⇒ 46.666 km/hr or 46.7 km/hr.
Hence the answer is = 46.666 km/hr or 46.7 km/hr.
Hope it helped u if yes mark me BRAINLIEST!
Tysm!
:)
Answer:
y = 43.55 + 2.15t
Explanation:
We were told that in 1983, the per capita consumption was 37.1 pounds, and in 1989 it was 50 pounds.
If we assume t = 0 corresponds to year 1980. Then, for 1983 it will be t = 3 and for 1989,it will be t = 9.
Thus, expressing the information as ordered pairs, we have; (3,37. 1) and (9,50).
Let us now find slope of the linear function:
m1 = (y2 - y1)/(t2 - t1)
m1 = (50 - 37.1)/(9 - 3)
m1 = 2.15
So, we can find the linear equation as;
y - 37.1 = 2.15(t - 3)
y = 37.1 + 2.15t - 6.45
y = 43.55 + 2.15t
Answer
given,
v = 128 ft/s
angle made with horizontal = 30°
now,
horizontal component of velocity
vx = v cos θ = 128 x cos 30° = 110.85 ft/s
vertical component of velocity
vy = v sin θ = 128 x sin 30° = 64 m/s
time taken to strike the ground
using equation of motion
v = u + at
0 =-64 -32 x t
t = 2 s
total time of flight is equal to
T = 2 t = 2 x 2 = 4 s
b) maximum height
using equation of motion
v² = u² + 2 a h
0 = 64² - 2 x 32 x h
64 h = 64²
h = 64 ft
c) range
R = v_x × time of flight
R = 110.85 × 4
R = 443.4 ft
