A 3.0 x 10^4 kg car is traveling to the East with a speed of 60 km/hr. Bigfoot jumps out onto the road causing the car to accelerate at 5.0 x 10^3 cm/s2 to the West. The car comes to a stop just in time for Bigfoot to capture and eat the driver of the car. How long does it take the car to stop from 60 km/hr to the East, and what was the displacement during this acceleration? Please give your answers in seconds and meters.
+37084 The car is travelling to the EAST at a constant rate (60km/hr) before it begins to DECCERATE to the WEST at a rate of 5 x 10^3 cm/s^2.
FIRST: The mass of the car is irrelevant....this is just 'smoke' to mislead you or see if you really know how to solve the problem.
SECOND: Get everything into the same units...lets use the decceleration units of cm/s^2
60 km/hr x 1000m/km x100cm/m / 3600 s/hr = 1666.667 cm/s
xf= xi + vt + 1/2 a t^2 we know v = 1666.667 cm/sec and a = -5 x 10^3 cm/s^2
We need to find t to solve this equation
We (should ) know vf= vfinal = 0 (because the car stops) = vi + a t If we substitute into THIS equation, we can find t to solve the FIRST equation and find xf
vf = 0 = 1666.667 + (-5 x 10^3) t yields t = .333 sec
Substitute this 't' into the first equation to solve for xf
xf = xo + vt + 1/2 a t^2
= 0 + 1666.667 (.333) + 1/2 (-5 x 10^3)(.333^2) = 277.77 cm = 2.778 meters
To stop from 60 km/hr in 2.778 meters is pretty impressive!!!
