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3 years ago

A train is traveling at 28 m/s. It slows to 6 m/s in 4 s. Calculate the distance the train

Physics

2 answers:

Eddi Din [679]3 years ago

8 0

Explanation:

A=v-u/t

A= 6- 28/4

A= 5.5m/s2

s= ut+ 1/2 at^2

s = 28 * 4+ 1/2 * 5.5 * 4^2

s = 112 + 1/2 * 5.5 * 16

s = 112 + 5.5* 8

s= 112 + 44

s = 156m

adell [148]3 years ago

7 0

The next step is -748 divide by -11 is 68 m (answer) the pic got cropped sorry

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James threw a ball vertically upward with a velocity of 41.67ms-1 and after 2 second David threw a ball vertically upward with a

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Answer:

When have passed 3.9[s], since James threw the ball.

Explanation:

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$y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\$

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$v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]$

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Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

$y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\$

As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

$y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]$

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

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