Question

A dog is walking right with a speed of 1.5 m/s sees a cat and speeds up with a constant rightward acceleration of magnitude 12 m/s^2. What is the velocity of the dog after speeding up for 3.0 m?

Answer 1

Answer:

8.617ms-¹ this is the velocity of the dog

Answer 2

Initial Speed of dog = 1.5 m/s

Acceleration attained by dog = 12 m/s²

• Find the velocity of dog after accelerating to 3 metres

We know,

$\bf{}S = ut + \frac{1}{2} {at}^{2} \\ \\ \{ \rm{putting \: \: the \: \: known \: \: values} \} \\ \\ \tt \to3 = 1.5(t) + \frac{1}{2} \times {12(t)}^{2} \\ \\ \tt \to3 = 1.5t + {6t}^{2} \\ \tt \to0 = 6 {t}^{2} + 1.5t - 3 \\ \\ \tt \to0 = {6t}^{2} + \frac{3}{2} t - 3 \\ \\ \tt \to0 = \frac{12 {t}^{2} + 3t - 6}{2} \\ \\ \tt \to 0 = {12 {t}^{2} + 3t - 6}^{} \\ \tt \to0 = 3( {4t}^{2} + t - 2) \\ \tt \to 0= {4t}^{2} + t - 2$

Solving the Quadratic Equation,

$= \rm {4t}^{2} + t - 2 \\ \\ \rm{}t = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ \rm t = \frac{ - 1 \pm \sqrt{ {1}^{2} - 4(4)( - 2)} }{2(4)} \\ \\ \rm t = \frac{ - 1 \pm \sqrt{1 + 32} }{8} \\ \\ \rm t = \frac{ - 1 \pm \sqrt{33} }{8} = \frac{ - 1 \pm5.74}{8} \\ \\ \rm t = \frac{ - 1 + 5.74}{8} \: \: \: \: or \: \: \: \: \frac{ - 1 - 5.74}{8} \\ \\ \rm t = \frac{4.74}{8} \: \: \: \: or \: \: \: \: \frac{ - 6.74}{8} \\ \\ \rm t = 0.5925 \: \: or \: \: - 0.8425$

Here, We were figuring out time, so the applicable value of t, can only be the positive result, because time can't be negative. {Also the unit of obtained result will be second as all the units which we calculated with were SI units}

Therefore,

• Time = 0.5925 sec

Now,

$\to \bf{v = u + at} \\ \\ \to \tt{v =1.5 + (12)(0.5925) } \\ \to \tt{v = 1.5 + 7.11} \\ \to \tt{v = \red{8.61} \: \: \rm {ms}^{ - 1} }$

Hence, It is concluded that,

• A dog walking with a speed of 1.5 m/s, when accelerates with a magnitude of 12 m/s², will have its final velocity as 8.61 m/s