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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
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B
40 m
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C
48 m
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D
46 m
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Solution

The correct option is A 52 m
The 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

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