Question

A car can travel at maximum speed of 180km/h and can have a maximum acceleration of 5m/s2 and retardation of 3m/s2. How fast can it start from rest and come to rest?​

Answer 1

Explanation:

ANSWER IS GIVEN BELOW

${v}^{2} - {u}^{2} = 2as \\ \\ = {50}^{2} - {0}^{2} = 2 \times 5 \times s \\ \\ = 2500 = 10s \\ \\ \frac{2500}{10} = s \\ \\ 250 = s \\ \\ now \\ \\ s = ut + \frac{1}{2} {at}^{2} \\ \\ = 250 = \frac{5}{2} {t}^{2} \\ \\ = \frac{250 \times 2}{5} = {t}^{2} \\ \\ = 100 = {t }^{2} \\ \\ \sqrt{100} = t \\ \\ 10 = t \\ \\ when \: retarding \\ \\ a = 3 \\ \\ t = \frac{v - u}{a} \\ \\ t = \frac{50 - 0}{3} \\ \\t = \frac{50}{3} \\ \\ t = 16.66677s$

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Answer 2

Answer:

Body will attain maximum speed in 10 sec and came to rest in 16.66 sec.

Explanation:

Given that :

• Maximum speed of car = 180 km/hrs
• Maximum acceleration = 5 m/sec²
• Maximum retardation = 3 m/sec²

To find :

• Time taken to attain maximum speed
• Time taken to come to rest

Solution :

• Let, the initial velocity of body be u, final velocity be v and acceleration of body be a1 and retardation of body be a2.
• Considering that the body starts from reat and accelerate with constant acceleration and attains a maximum speed of 180 km/hrs.
• Initial velocity, u = 0 m/sec
• Final velocity, v = 180 km/hrs = 50 m/sec
• acceleration, a1 = 5 m/sec²
• From 1st equation of motion,

$v = u + at \\ 50 = 5(t) \\ t = 10 \: sec$

• Hence, body will attain maximum speed in 10 sec.
• Again, considering that body retards with constant retardation, a2 = 3 m/sec² and came to rest.
• Initial velocity, v = 50 m/sec
• Final velocity, v = 0 m/sec
• retardation, a2 = -3 m/sec²
• From 1st equation of motion,

$v = u + at \\ 0 = 50 - 3t \\ t = 16.6 \: sec$

• Hence, body will take 16.66 sec to came to rest.