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A car travelling with uniform acceleration covers 65m in the 5th second and 115m in the 10th second of its motion. Calculate the distance travelled by the car (i) in the 7th second of its motion and (ii) 7 seconds of its motion



Answer :

lexie2fine

Answer:  Since the car is traveling with uniform acceleration, we can use the equation of motion:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

We're given two pieces of information:

The car covers 65m in the 5th second of its motion.

The car covers 115m in the 10th second of its motion.

Let's use these to find the acceleration (a) and initial velocity (u).

First, let's find the velocity at the 5th second:

v_5 = distance / time = 65m / 5s = 13m/s

Now, let's find the velocity at the 10th second:

v_10 = distance / time = 115m / 10s = 11.5m/s

Since the acceleration is uniform, we can use the equation:

v_10 = v_5 + a(10 - 5)

Substituting the values, we get:

11.5m/s = 13m/s + a(5)

Simplifying, we get:

a = -0.5m/s²

Now that we have the acceleration, let's find the initial velocity (u). We can use the equation:

v_5 = u + a(5)

Substituting the values, we get:

13m/s = u - 0.5m/s²(5)

Simplifying, we get:

u = 18m/s

Now, let's answer the questions:

(i) Distance traveled in the 7th second:

We can use the equation:

distance = u × t + (1/2) × a × t²

Substituting the values, we get:

distance = 18m/s × 7s + (1/2) × (-0.5m/s²) × (7s)²

distance ≈ 94.5m

(ii) Distance traveled in 7 seconds:

We can use the equation:

distance = u × t + (1/2) × a × t²

Substituting the values, we get:

distance = 18m/s × 7s + (1/2) × (-0.5m/s²) × (7s)²

distance ≈ 343.5m

So, the answers are:

(i) The distance traveled by the car in the 7th second of its motion is approximately 94.5m.

(ii) The distance traveled by the car in 7 seconds of its motion is approximately 343.5m.

Explanation:  

MathPhys

Answer:

(i) 85 m

(ii) 385 m

Explanation:

The car moves with uniform acceleration, so its motion can be modeled using constant acceleration kinematics, also known as SUVAT. Specifically, we will use the equation:

s = ut + ½ at²

where

  • s is the displacement (the position of the car from its starting point)
  • u is the initial velocity (the initial speed and direction)
  • a is the acceleration
  • t is time

First, let's use this equation to write expressions for the car's position at the end of 4 seconds, 5 seconds, 9 seconds, and 10 seconds.

s = ut + ½ at²

s₄ = u(4) + ½ a(4)² = 4u + 8a

s₅ = u(5) + ½ a(5)² = 5u + 12.5a

s₉ = u(9) + ½ a(9)² = 9u + 40.5a

s₁₀ = u(10) + ½ a(10)² = 10u + 50a

Next, we know the difference in position between 4 and 5 seconds is 65 meters.

s₅ − s₄ = 65

(5u + 12.5a) − (4u + 8a) = 65

u + 4.5 a = 65

And we know the difference in position between 9 and 10 seconds is 115 meters.

s₁₀ − s₉ = 115

(10u + 50a) − (9u + 40.5a) = 115

u + 9.5 a = 115

Solve the system of equations. Subtracting the first equation from the second:

5a = 50

a = 10

u = 20

Therefore, the car's position at time t is:

s = 20t + 5t²

We can use this equation to solve (i) and (ii).

(i) In the 7th second, the car travels a distance of:

s₇ − s₆ = [20(7) + 5(7)²] − [20(6) + 5(6)²]

s₇ − s₆ = 85

(ii) At the end of 7 seconds, the car has travelled:

s₇ = 20(7) + 5(7)²

s₇ = 385

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