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givi [52]
3 years ago
11

Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

oment the cars are 45 m apart, the lead driver applies the brakes, causing the car to have an acceleration of â2.0 m/s².
a. How long does it take for the lead car to stop?
b. Assume that the driver of the chasing car applies the brakes at the same time as the driver of the lead car. What must the chasing carâs minimum negative acceleration be to avoid hitting the lead car?
c. How long does it take the chasing car to stop?
Physics
1 answer:
Phoenix [80]3 years ago
7 0

Answer:

a. t_1=12.5\ s

b. a_2=-13.61\ m.s^{-2}  must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. t_2=2.5714\ s is the time taken to stop after braking

Explanation:

Given:

  • speed of leading car, u_1=25\ m.s^{-1}
  • speed of lagging car, u_{2}=35\ m.s^{-1}
  • distance between the cars, \Delta s=45\ m
  • deceleration of the leading car after braking, a_1=-2\ m.s^{-2}

a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

where:

v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

where:

v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

0^2=35^2+2\times a_2\times 45

a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

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Let's substitute the value of v

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Let's calculate

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    V₁₀ = 7.668 (2.68) / 10.82

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