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svetoff [14.1K]
3 years ago
10

Two cars start from rest and begin accelerating, one at 10m/s210m/s2 and the other at 12m/s212m/s2. How long will it take for th

e second car to be 100m100m ahead of the first car? A :
Physics
1 answer:
Vaselesa [24]3 years ago
3 0

Answer:

The time taken by the second car to be 100 m ahead of the first car is 10 seconds.

Explanation:

Given that,

Initial speeds of both car is 0 as they starts at rest.

The acceleration of car 1, a_1=10\ m/s^2

The acceleration of car 2, a_2=12\ m/s^2

The distance covered by car 1 is :

d_1=ut+\dfrac{1}{2}a_1t^2

d_1=\dfrac{1}{2}at^2

d_1=\dfrac{1}{2}\times 10t^2

d_1=5t^2.......(1)

The distance covered by car 2 is :

d_2=ut+\dfrac{1}{2}a_2t^2

d_2=\dfrac{1}{2}a_2t^2

d_2=\dfrac{1}{2}\times 12t^2

d_2=6t^2......(2)

It is mentioned that the second car to be 100 m ahead of the first car, so,

6t^2-5t^2=100

t^2=100

t = 10 seconds

So, the time taken by the second car to be 100 m ahead of the first car is 10 seconds. Hence, this is the required solution.

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a) v₁fin = 3.7059 m/s   (→)

b) v₂fin = 1.0588 m/s     (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s   ⇒  Kin = 0 J

v₁fin = ?

h<em>in </em>= L = 0.7 m

h<em>fin </em>= 0 m   ⇒    U<em>fin</em> = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

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⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s   (→)

b)  Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s    (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒  v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒  v₂fin = 1.0588 m/s     (→)

7 0
3 years ago
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