Question

# A car starts from rest and accelerates at  for 4s. It then moves at a constant velocity for next $$1 m{ s }^{ -2}$$ , then decelerates at $$2.0 m{ s }^{ -2}$$ and finally comes to rest . The total distance traveled by the car is

A
52 m
B
40 m
C
48 m
D
46 m

Solution

## The correct option is A 52 mThe distance travelled by the car in first 4 seconds: $$x_{1}=ut+\frac{1}{2}at^{2}$$$$x_{1}=0(t)+\frac{1}{2}\times 1\times (4)^{2}$$$$x_{1}=8m......Eq:01$$The velocity (v) at the end of these 4 seconds with constant velocity is:$$v=u+at$$$$v=0+(1)(4)$$$$v=4ms^{-1}$$New distance $$x_{2}$$ in the next 10 seconds will be:$$x_{2}=v\times 10$$$$x_{2}=4\times 10=40m......Eq:02$$Use formula to find the distance when the car started retarded:$$v^{2}-u^{2}=2a(x_{3})$$$$(4)^{2}-0^{2}=2(2)\times (x_{3})$$$$16=4\times(x_{3})$$$$(x_{3})=4m......Eq:03$$Required total distance travelled:Adding Eq:01; Eq:02 and Eq:03;$$d=x_{1}+x_{2}+x_{3}$$$$d=(8+40+4)m$$$$d=52m$$OPTION (A) IS CORRECT.Physics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More