Question

a child operating a radio-controlled model car on a dock accidentally steers it off the edge. the car's displacement 0.70 s after leaving the dock has a magnitude of 8.0 m. what is the car's speed at the instant it drives off the edge of the dock?

Answer 1

`v_(0x) = 10.90\ m/s`

Step-by-step explanation:

given,

time = 0.7 s

displacement of the car = r = 8 m

vertical displacement of the car

using equation of motion

`y= ut + (1)/(2)gt^2`

initial velocity of the car is equal to zero

`y=(1)/(2)gt^2`

now, horizontal distance traveled by the car

r² = x² + y²

`x = √(r^2-y^2)`

`v_(0x) =(x)/(t)`

`v_(0x) =(√(r^2-y^2))/(t)`

`v_(0x) =\frac{\sqrt{r^2-((1)/(2)gt^2)^2}}{t}`

`v_(0x) =\frac{\sqrt{8^2-((1)/(2)(-9.8)* 0.7^2)^2}}{0.7}`

`v_(0x) = 10.90\ m/s`