The distance x metres travels by a vehicle in time t seconds after the brakes are applied is given by x=20t-5÷3t^2,determine a the
1)speed of the vehicle in km/h at the instance the brake are apply
2)the distance the car travels before it stops
To find the speed of the vehicle in km/h at the instance the brakes are applied, we need to differentiate the equation for distance with respect to time and then convert the units.
1) Speed of the vehicle in km/h:
Differentiating the equation x = 20t - (5/3)t^2 with respect to t gives:
dx/dt = 20 - (10/3)t
To convert the units to km/h, we divide by 1000 to convert meters to kilometers, and multiply by 3600 to convert seconds to hours:
dx/dt (in km/h) = (20 - (10/3)t) * (1/1000) * 3600
Substituting t = 0 (the instant the brakes are applied) into the equation gives:
dx/dt (at t = 0) = (20 - (10/3)(0)) * (1/1000) * 3600
dx/dt (at t = 0) = 20 * (1/1000) * 3600
dx/dt (at t = 0) = 72 km/h
Therefore, the speed of the vehicle in km/h at the instance the brakes are applied is 72 km/h.
2) The distance the car travels before it stops:
The car will stop when the velocity becomes zero. To find this point, we set dx/dt = 0 and solve for t:
dx/dt = 20 - (10/3)t = 0
20 = (10/3)t
60 = 10t
t = 6 seconds
Substituting this value of t back into the equation for distance gives:
x = 20(6) - (5/3)(6)^2
x = 120 - (5/3)(36)
x = 120 - 60
x = 60 meters
Therefore, the distance the car travels before it stops is 60 meters.