Answer:
V = (5.8cm/s)i, (4.7cm/s)j
Explanation:
Given :
r⃗ =[ 4.50 cm +( 2.90 cm/s2 )t2]i^+( 4.70 cm/s )tj^
To obtain the average velocity (V)
V = (r2 - r1) / (t2 - t1)
To obtain r1 and r2, substitute t1 = 0 and t2 = 2 respectively in the equation above
r1 = [ 4.50 cm +( 2.90 cm/s2 ) 0]i^+( 4.70 cm/s )0 j
r1 = 4.50 cm + 0 + 0 = (4.50cm)i + 0j
r2 = [ 4.50 cm +( 2.90 cm/s2 )2²]i^+( 4.70 cm/s )2 j
r2 = 4.50cm + (2.90 × 4)i + (4.70 × 2)j
r2 = (16.1cm)i + (9.4cm)j
V = [(16.1 - 4.50)i - (9.4 - 0)j] / 2 - 0
V = 11.6i / 2 ; 9.4j / 2
V = (5.8cm/s)i, (4.7cm/s)j
Average speed = total distance / total time
= (81 + 88) / (2 + 2)
= 169/4
= 42.25 kilometers/hour
Let me know if you have any questions.
Answer:
dR/dt = 10.2 ft / s
Explanation:
Let's work this problem by finding the distance between the balloon and the motorcycle and then drift for the speed change of the distance
Balloon
y = y₀ +
t
Motorcycle
x = v₀ₓ t
Distance, let's use Pythagoras' theorem
R² = x² + y²
R² = (v₀ₓ t)² + (y₀ + t)²
v₀ₓ = 88 ft / s
= 8 ft / s
y₀ = 150 ft
R² = (8 t)² + (150 + 8 t )²
R² = 64 t² + (150 + 8t )²
This is the expression for the distance between the two bodies, the rate of change is the derivative with respect to time (d / dt)
2RdR / dt = 64 2 t + 2 (150 + 8t) 8
dR / dt = [64 t + (1200 + 64t )] / R
dR/dt = (1200 +128 t)/R
Let's calculate for the time of 10 s
dR / dt = (1200 + 128 10) / R = 2480 /R
R = √ [64 10² + (150 + 8 10)²
R = √ [6400 + 52900]
R = 243.5 ft
dR / dt = (2480) / 243.5
dR / dt = 10.2 ft / s
Answer:
B. Longer than t s,
Explanation:
Gravitational accln on earth is 9.8 m/s^2,
and one you provided as on moon is 1.6 m/s^2
that mean on moon gr. accl. is lesser!
now the time taken on earth will be lesser cuz from the same height if you drop the object from rest!
since accln on earth is higher,the object will attain higher velocity as compare to that of on moon!
✌️:)
