Physics, asked by manokingpuli907, 1 month ago

7. A car is crossing a straight long bridge. The car maintains a uniform acceleration. If it has a velocity of 2 m/s when the crossing starts and a velocity of 14 m/s when the crossing is finished, the velocity of the car when it has covered half of the length of the bridge is a. 8m/s b.12 m/s c. 10 m/s d. 9m/s​

Answers

Answered by jagruti6551
2

Answer:

If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the shore, as in Figure 1. The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes see that the plane is not moving in the direction in which it is pointed, as illustrated in Figure 2. The plane is moving straight ahead relative to the air, but the movement of the air mass relative to the ground carries it sideways

Answered by rishkrith123
0

Answer:

The velocity of the car when it has covered half of the length of the bridge is 10 m/s.

Explanation:

Given,

The initial velocity of the car is (u) = 2m/s

The final velocity of the car is (v) = 14 m/s

To find,

The velocity of the car when it has covered half of the length of the bridge(V).

Assumption,

Let the total length of the straight long bridge be 'S'

Let the uniform acceleration with which the car is traveling be 'a'

Calculation,

From equations of motion, we know that

v² - u² = 2as...(1)

Here v = 14 m/s, u = 2 m/s, s = S, a = a

⇒ 14² - 2² = 2aS

⇒ 196 - 4 = 2aS

⇒ aS = 96...(2)

Now for the velocity of the car when it has covered half of the length of the bridge is:

v = V, u = 2m/s, s = S/2, a = a

Substituting in equation (1) we get:

V² - 2² = 2a(S/2)

⇒ V² - 4 = aS

From equation (2) we get:

V² - 4 = 96

⇒ V² = 100

⇒ V = 10 m/s

Therefore, the velocity of the car when it has covered half of the length of the bridge is 10 m/s.

#SPJ3

Similar questions