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Grade upto college level Electric Current

at time t=0 a car moving with velocity of 16ms. it slows down with an acceleration of -0.5t where t is in second. tick the correct

a) the direction of velocity changes at= 8 seconds

b) the distance travelled in 4 seconds is approx.58.67m

c) distance travelled in 10 s is 94m

d)the speed at t=10seconds is 9ms

Profile image of Kevin Nash
12 Years agoGrade upto college level
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1 Answer

Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer1 Year ago

To tackle this problem, we need to analyze the motion of the car given its initial velocity and the acceleration that varies with time. The car starts with a velocity of 16 m/s and experiences a deceleration of -0.5t m/s². Let's break down each option step by step to determine which statements are correct.

Understanding the Motion

The car's acceleration is not constant; it changes with time. At any time t, the acceleration can be expressed as:

a(t) = -0.5t

This means that the acceleration decreases as time increases, leading to a change in velocity over time. We can find the velocity at any time t by integrating the acceleration.

Finding the Velocity Function

To find the velocity as a function of time, we integrate the acceleration:

v(t) = v(0) + ∫a(t) dt

Given that the initial velocity v(0) is 16 m/s, we have:

v(t) = 16 + ∫(-0.5t) dt

Calculating the integral:

v(t) = 16 - 0.25t² + C

Since the initial velocity at t=0 is 16 m/s, we find that C = 0. Thus:

v(t) = 16 - 0.25t²

Evaluating Each Statement

  • a) The direction of velocity changes at t = 8 seconds: To determine when the velocity changes direction, we set v(t) to zero:
  • 0 = 16 - 0.25t²

    0.25t² = 16

    t² = 64

    t = 8 seconds

    This statement is correct; the velocity changes direction at t = 8 seconds.

  • b) The distance travelled in 4 seconds is approximately 58.67 m: To find the distance, we need to integrate the velocity function:
  • s(t) = ∫v(t) dt = ∫(16 - 0.25t²) dt

    s(t) = 16t - (0.25/3)t³ + C

    At t = 0, s(0) = 0, so C = 0:

    s(t) = 16t - (1/12)t³

    Now, substituting t = 4 seconds:

    s(4) = 16(4) - (1/12)(4)³ = 64 - (1/12)(64) = 64 - 5.33 ≈ 58.67 m

    This statement is also correct.

  • c) Distance travelled in 10 seconds is 94 m: Let's calculate s(10):
  • s(10) = 16(10) - (1/12)(10)³ = 160 - (1/12)(1000) = 160 - 83.33 ≈ 76.67 m

    This statement is incorrect.

  • d) The speed at t = 10 seconds is 9 m/s: We can find v(10):
  • v(10) = 16 - 0.25(10)² = 16 - 25 = -9 m/s

    The speed is 9 m/s, but since it's negative, it indicates the car is moving in the opposite direction. Therefore, this statement is also correct.

Summary of Findings

Based on our calculations:

  • a) Correct
  • b) Correct
  • c) Incorrect
  • d) Correct (but note the direction)

In conclusion, options a, b, and d are correct, while option c is incorrect. Understanding how to integrate acceleration to find velocity and then distance is crucial in solving such problems. If you have any further questions or need clarification on any part, feel free to ask!