Motion Analysis Under Changing Acceleration

The problem asks us to find the value of time t for a vehicle that starts from rest and moves in two distinct phases with different constant accelerations, covering a total distance of 550 m. We need to apply the equations of motion for constant acceleration.

Phase 1: Initial Motion

In the first phase, the vehicle starts from rest and moves with a constant acceleration for time t.

• Initial velocity (\(u_1\)): \(0 \ m/s\) (since it starts from rest)
• Acceleration (\(a_1\)): \(2 \ m/s^2\)
• Time (\(t_1\)): t seconds

We can find the distance traveled (\(s_1\)) during this phase using the kinematic equation: \(s = ut + \frac{1}{2}at^2\)

Substituting the values for phase 1: \(s_1 = (0)(t) + \frac{1}{2}(2 \ m/s^2)(t^2)\) \(s_1 = t^2 \ m\)

We also need the velocity (\(v_1\)) at the end of this phase, as it will be the initial velocity for the next phase. Using the equation v = u + at: \(v_1 = u_1 + a_1 t_1\) \(v_1 = 0 + (2 \ m/s^2)(t)\) \(v_1 = 2t \ m/s\)

Phase 2: Subsequent Motion

In the second phase, the vehicle moves with a different acceleration for a duration of 10 seconds.

• Initial velocity (\(u_2\)): \(v_1 = 2t \ m/s\)
• Acceleration (\(a_2\)): \(5 \ m/s^2\)
• Time (\(t_2\)): \(10 \ s\)

The distance traveled (\(s_2\)) during this phase is calculated using \(s = ut + \frac{1}{2}at^2\): \(s_2 = u_2 t_2 + \frac{1}{2} a_2 t_2^2\) \(s_2 = (2t \ m/s)(10 \ s) + \frac{1}{2}(5 \ m/s^2)(10 \ s)^2\) \(s_2 = 20t \ m + \frac{1}{2}(5 \ m/s^2)(100 \ s^2)\) \(s_2 = 20t \ m + (5)(50) \ m\) \(s_2 = (20t + 250) \ m\)

Total Distance Calculation

The total distance traveled is the sum of the distances covered in both phases (\(s_{total} = s_1 + s_2\)). We are given that the total distance is 550 m.

Therefore: \(s_1 + s_2 = 550 \ m\) \(t^2 + (20t + 250) = 550\)

Solving the Quadratic Equation for Time t

Now, we need to solve the equation for t: \(t^2 + 20t + 250 = 550\)

Subtract 550 from both sides to set the equation to zero: \(t^2 + 20t + 250 - 550 = 0\) \(t^2 + 20t - 300 = 0\)

This is a quadratic equation in the form \(at^2 + bt + c = 0\), where a=1, b=20, and c=-300. We can solve for t using the quadratic formula: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Substitute the values of a, b, and c: \(t = \frac{-20 \pm \sqrt{(20)^2 - 4(1)(-300)}}{2(1)}\) \(t = \frac{-20 \pm \sqrt{400 + 1200}}{2}\) \(t = \frac{-20 \pm \sqrt{1600}}{2}\) \(t = \frac{-20 \pm 40}{2}\)

This gives two possible values for t:

1. \(t = \frac{-20 + 40}{2} = \frac{20}{2} = 10\)
2. \(t = \frac{-20 - 40}{2} = \frac{-60}{2} = -30\)

Since time cannot be negative in this physical context, we discard the negative value.

Therefore, the value of time t is \(10 \ s\).

Conclusion

The calculated value of t matches option 1.