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MCQEasyJEE 2025Velocity & Acceleration

JEE Physics 2025 Question with Solution

A person travelling on a straight line moves with a uniform velocity v1v_1 for a distance xx and with a uniform velocity v2v_2 for the next 3x2\frac{3x}{2} distance. The average velocity in this motion is 507 m/s\frac{50}{7} \, \text{m/s}. If v1v_1 is 5 m/s5 \, \text{m/s}, then v2v_2 is _____ m/s\text{m/s}.

  • A

    10 m/s10 \, \text{m/s}

  • B

    12 m/s12 \, \text{m/s}

  • C

    15 m/s15 \, \text{m/s}

  • D

    18 m/s18 \, \text{m/s}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The person travels distance xx with velocity v1=5 m/sv_1 = 5 \, \text{m/s} and distance 3x2\frac{3x}{2} with velocity v2v_2. The average velocity is 507 m/s\frac{50}{7} \, \text{m/s}.

Find: The value of v2v_2.

For motion along a straight line without change of direction, average velocity equals total displacement divided by total time.

vavg=x1+x2t1+t2v_{\text{avg}} = \frac{x_1 + x_2}{t_1 + t_2}

Here,

x1=x,x2=3x2x_1 = x, \qquad x_2 = \frac{3x}{2}

So total distance is

Dtotal=x+3x2=5x2D_{\text{total}} = x + \frac{3x}{2} = \frac{5x}{2}

Time taken in the two parts is

t1=xv1,t2=3x2v2t_1 = \frac{x}{v_1}, \qquad t_2 = \frac{3x}{2v_2}

Hence,

507=5x2x5+3x2v2\frac{50}{7} = \frac{\frac{5x}{2}}{\frac{x}{5} + \frac{3x}{2v_2}}

Factor out xx from the denominator:

507=5215+32v2\frac{50}{7} = \frac{\frac{5}{2}}{\frac{1}{5} + \frac{3}{2v_2}}

Simplifying,

507=5v1v23v1+2v2\frac{50}{7} = \frac{5v_1v_2}{3v_1 + 2v_2}

Substitute v1=5v_1 = 5:

507=25v215+2v2\frac{50}{7} = \frac{25v_2}{15 + 2v_2}

Cross-multiplying,

507(15+2v2)=25v2\frac{50}{7}(15 + 2v_2) = 25v_22(15+2v2)=7v22(15 + 2v_2) = 7v_230+4v2=7v230 + 4v_2 = 7v_230=3v230 = 3v_2v2=10v_2 = 10

Therefore, v2=10 m/sv_2 = 10 \, \text{m/s} and the correct option is A.

Step-by-Step Distance-Time Method

Given: Average velocity =507 m/s= \frac{50}{7} \, \text{m/s}, first velocity v1=5 m/sv_1 = 5 \, \text{m/s}, first distance =x= x, second distance =3x2= \frac{3x}{2}.

Find: The second velocity v2v_2.

Using

vavg=Total DistanceTotal Timev_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}}

Total distance:

Dtotal=x+3x2=5x2D_{\text{total}} = x + \frac{3x}{2} = \frac{5x}{2}

Total time:

Ttotal=xv1+3x2v2T_{\text{total}} = \frac{x}{v_1} + \frac{3x}{2v_2}

Therefore,

vavg=5x2xv1+3x2v2v_{\text{avg}} = \frac{\frac{5x}{2}}{\frac{x}{v_1} + \frac{3x}{2v_2}}

Substitute vavg=507v_{\text{avg}} = \frac{50}{7} and v1=5v_1 = 5:

507=5x2x5+3x2v2\frac{50}{7} = \frac{\frac{5x}{2}}{\frac{x}{5} + \frac{3x}{2v_2}}

Cancel xx:

507=5215+32v2\frac{50}{7} = \frac{\frac{5}{2}}{\frac{1}{5} + \frac{3}{2v_2}}

Take the denominator in a single fraction:

15+32v2=2v2+1510v2\frac{1}{5} + \frac{3}{2v_2} = \frac{2v_2 + 15}{10v_2}

So,

507=52×10v22v2+15\frac{50}{7} = \frac{5}{2} \times \frac{10v_2}{2v_2 + 15}507=25v22v2+15\frac{50}{7} = \frac{25v_2}{2v_2 + 15}

Cross-multiply:

50(2v2+15)=175v250(2v_2 + 15) = 175v_2100v2+750=175v2100v_2 + 750 = 175v_2750=75v2750 = 75v_2v2=10v_2 = 10

Therefore, the value of v2v_2 is 10 m/s10 \, \text{m/s}.

Common mistakes

  • Using the arithmetic mean v1+v22\frac{v_1 + v_2}{2} for average velocity is incorrect because the distances covered in the two segments are unequal. Use total distance divided by total time instead.

  • Adding distances correctly but not computing the separate times for each segment leads to a wrong equation. First find t1=xv1t_1 = \frac{x}{v_1} and t2=3x2v2t_2 = \frac{3x}{2v_2}, then use vavg=DtotalTtotalv_{\text{avg}} = \frac{D_{\text{total}}}{T_{\text{total}}}.

  • Cancelling xx too early without factoring it from the denominator can produce algebraic errors. Write the denominator as x(1v1+32v2)x\left(\frac{1}{v_1} + \frac{3}{2v_2}\right) and then cancel xx safely.

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