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A proton travelling with a velocity of 3×105,2×105,-4×105m/spasses the origin at a time 9.0 s after a proton detector is turned on. Assuming that the velocity of the proton does not change, what will be its position at time 9.7 s?

Short Answer

Expert verified

At time 9.7 s , the proton will be present at 2.1×105,1.4×105,-2.8×105m .

Step by step solution

01

Identification of a given information

The given information can be listed as follows:

  • The initial position vector is,r→i=0,0,0m .
  • The average velocity is,V→avg=3×105,2×105,-4×105m/s .
  • The initial time is,ti=9.0s .
  • The final time is,tf=9.7s .
02

Definition of the average velocity

Average velocity is the ratio of an object’s displacement and time taken.

It can be expressed as follows:

V→avg=r→f-r→itf-ti

Here,r→fandr→iare final and initial position vectors;and are final tfand tithe initial time.

From the formula of average velocity, r→fcan be derived as follows:

role="math" localid="1653479554793" r→f=r→i+V→avgtf-ti.

03

Determination of the final position

Substitute the value of r→i, V→avg, ti, and into the equation role="math" localid="1653480484382" r→f=r→i+V→avgtf-ti.

role="math" localid="1653480553527" r→f=0,0,0m+3×105,2×105,-4×1059.7-9.0m=0,0,0m+3×105,2×105,-4×1050.7m=2.1×105,1.4×105,-2.8×105m

Hence, the final position of the proton is2.1×105,1.4×105,-2.8×105m.

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