Question

The instantaneous height y and the horizontal distance x covered by a particle are as follows : 

y = bt²., x = ct². What is the speed of the particle one second after the firing ? 

(a) 2(b + c) 

(b) 2(b - c) 

(c) 2(b² + c)^1/2 

(d) 2(b² - c²)^1/2. 

Answer 1

Answer is (c) 2(b² + c)^1/2

νx = 2ct and νy = 2bt.. speed after one second will be

[(2b)² + (2c)²]^1/2 = 2(b² + c²)^1/2