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Anonymous asked in Science & MathematicsPhysics · 8 months ago

A car traveling 67 km/h slows down at a constant 0.45 m/s2 just by "letting up on the gas."?

Update:

Calculate the distance the car coasts before it stops. (Answer 380 m)

Calculate the time it takes to stop. (Answer 41s)

Calculate the distance it travels during the first second. (Answer 18 m)

Calculate the distance it travels during the fifth second. I JUST NEED HELP WITH PART D

4 Answers

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  • oubaas
    Lv 7
    8 months ago

    A car traveling 67 km/h slows down at a constant 0.45 m/s2 just by "letting up on the gas."

    1) Calculate the distance d the car coasts before it stops to two significant figures and include the appropriate units.

    d =V^2/2a = 67^2/(3.6^2*0,9) = 380 m

    2) Calculate the time t it takes to stop to two significant figures and include the appropriate units.

    t = V/a = 67/(3.6*0.45) = 41 sec

    3) Calculate the distance d1 it travels during the first second to two significant figures

    d1 = V*1-a/2*1^2 = 67/3.6-0.225 = 18 m

    d1 = (67/3.6+67/3.6-0.45*1)/2*1 = 18 m

    4) Calculate the distance d2 it travels during the fifth second to two significant figures

    V4th = (67/3.6-0.45*4) = 16.81 m/sec

    V5th = (67/3.6-0.45*5) = 16.36 m/sec

    d2 = (16.81+16.36)/2*1 = 17.0 m

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  • 8 months ago

    Step 1 convert the speed to SI 67 km / hr = 67000m/ 3600 s

    ~= 18.6 m/s

    now for a) v^2 = 2as ->s = v^2 / ( 2a) = 18.61^2 / 0.90 = 385 m

    b) v = at -> t = v/a = 18.61/ 0.45 = 41 s

    c) distance in the first second = (v(0)+v(1))/2 = ( v + v-0.45) / 2 = v- 0.45/2 = 18.4 m

    d) distance in the fifth second =( v(4)+ v(5)) / 2 * 1

    = ( v- 4*0.45 + (v - 5*0.45) ) / 2 * 1

    = v - 9*0.45/2 ~= 16.6 m

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  • 8 months ago

    The velocity is changing linearly (constant acceleration) so the average velocity over any time interval can be calculated as either of:

    .... the average of the starting and ending velocities, or

    .... the velocity at the midpoint (in *time*, not distance!).

    So the average velocity over [4 s, 5 s] is the speed at t=4.5 s:

    v = 67 km/h - (0.45 m/s^2)(4.5 s)

    = 67 km/h * [1000 m / 1 km] * [1 h / 3600 s] + 2.025 m/s

    = 18.61 m/s - 2.03 m/s

    = 20.64 m/s

    That's too many digits, but I like to round at the very end. In one second at that average speed you travel about 20.64 meters, rounding to 21.6 assuming the 67 km/h initial speed was within about 0.5 m/s which is about 0.14 m/s; which I think is small enough to quote the first decimal place. The delta-v figure has 2 significant digits, so 2.03 m/s is good to the first decimal place, too.

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  • Anonymous
    8 months ago

    67 km/hr = 18.61 m/sec

    At the end of the 4th second (start of the 5th second), the new speed = 18.61 m/sec - 0.45 m/sec^2 * 4 sec = 16.81 m/sec. At the end of the 5th second, the vehicle is going 16.81 m/sec - 0.45 m/sec^2 * 1 sec = 16.36 m/sec

    vf^2 = vi^2 + 2 * a * d

    Substitute and solve for d.

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