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vodomira [7]
3 years ago
9

Pascal has 96 miles remaining to complete his cycling trip. If he reduced his current speed by 4 miles per hour, the remainder o

f the trip would take him 16 hours longer than it would if he increased his speed by 50%. What is his current speed?
Physics
1 answer:
Verdich [7]3 years ago
6 0

Answer:

V = 20 miles /sec

Explanation:

We have remaining distance   =  d  = 96 miles

Lets call  Pascal velocity  V in miles per hour

Now if he increases his velocity by  50 % (equivalent to multiply by 1.5 ) he will need a time t₁ to arrive then as V = d/t

1.5* V  = d/ t₁      ⇒   1.5 * V  =  96 /t₁

And in the case of reducing his velocity

(V / 4) = d/ (t₁ + 16 )     ⇒  V * (t₁ + 16 ) = 4*d     ⇒ V*t₁ + 16*V = 384

So we a 2 equation system with two uknown variables

1.5*V = 96/t₁      (1)

V*t₁  + 16*V = 384     (2)

We solve  from equation    (1)      t₁  = 64/V

And by substitution   in equation (2)

V * (64/V) + 16* V = 384

64  + 16 *V  = 384         ⇒   16*V = 320      ⇒  V= 320/16

V = 20 miles /sec

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Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri
Ainat [17]

Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

  The speed of car A is  v_A  =  22 \ m/s

   The speed of car B is  v_B  = 29.0 \ m/s

     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is  

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Where t_B is the time taken by car B

Now this can also be represented as using equation of motion as

      d = v_A t_A  + \frac{1}{2}a_A t_A^2 - 300

Now substituting values

       d = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

Equating the both d

       v_B * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

substituting values

   29 * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

  7 t_A =1.2 t_A^2 - 300

   1.2 t_A^2 - 7 t_A - 300  = 0

Solving this using quadratic formula we have that

     t_A  = 19.0 \ s

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