What is the speed of the car when it passes the truck?
A car moving at a velocity of 20.0 m/s is behind a truck moving at a constant velocity of 18.0 m/s. When the car is 50.0 m behind the front of the truck, the car accelerates uniformly at 1.80 m/s^2. The car continues at the same acceleration until it passes the front of the truck.
A car moving at a velocity of 20.0 m/s is behind a truck moving at a constant velocity of 18.0 m/s. When the car is 50.0 m behind the front of the truck, the car accelerates uniformly at 1.80 m/s^2. The car continues at the same acceleration until it passes the front of the truck.
1 Answer
Let the car pass the truck after elapse of time
- Movement of truck.
It continues at a constant velocity of#18.0\ ms^-1#
Distance covered by truck in time#t=18.0t\ m# - Distance covered by car during time
#t# . Applicable kinematic expression is#s=ut+1/2at^2#
Inserting various values we get
#s=20.0t+1/2xx1.80t^2#
#=>s=20.0t+0.90t^2# - For the car to overtake truck in time
#t# it must cover distance equal to the sum of distance traveled by truck and#50.0\ m# . Therefore we have#s=18.0t+50.0#
Equating#s# calculated in steps 2. and 3. we get
#20.0t+0.90t^2=18.0t+50.0#
#=>0.9t^2+2t-50=0# - Solving the quadratic graphically (we can use analytical method as well) we get
#t=-8.647and 6.425#
Ignoring#-ve# root as time can not be negative we get
#t=6.425\ s#
Speed of the car when it passes truck is found from the kinematic expression
#v=u+at#
#v=20.0+1.80xx6.425#
#v=31.56\ ms^-1#
