Question

A car starting from rest accelerates at the rate f through a distance s than continuous at constant speed for time t and then then decelerates at the rate f/2 come to rest if the total distance travelled is 15 second then

Answer 1

May u like the solution

Answer 2

Answer:

Value of s is ft²/72 .

Explanation:

Given that :

• 1st part : Car starts from rest and accelerate at rate f through a distance s
• 2nd part : Car travels at continuous speed for time t
• 3rd part : Car decelerates at the rate f/2 and came to rest
• Total distance travelled = 15s

Solution :

• For 1st part, the final velocity of car is given by 3rd equation of motion

${v}^{2} - {u}^{2} = 2as$

• initial velocity, u1 = 0
• acceleration, a1 = f
• distance travelled, s1 = s

${(v1)}^{2} - {(u1)}^{2} = 2(a1)(s1) \\ {(v1)}^{2} = 2 fs \\ v1 = \sqrt{2fs}$

• For 2nd part, car moves with this constant velocity for time t
• distance travelled, s2 = ( v1 x t )

$s2 = ( \sqrt{2fs} \: ) t$

• For 3rd part, car decelerates and came to rest
• initial velocity, v1 = (√2fs)
• final velocity, v3 = 0
• acceleration, a3 = f/2
• Let, distance travelled = s3

${v3}^{2} - {v1}^{2} = 2(a3)(s3) \\ - {v1}^{2} = - 2( \frac{f}{2} )(s3) \\ 2fs= f(s3) \\ (s3) = 2s$

• Total distance travelled by car is 15s.

$15s = s1 + s2 + s3 \\ 15s = s + (\sqrt{2fs} )t + 2s \\ 12s = (\sqrt{2fs} ) t \\ 144 {s}^{2} = 2fs {t}^{2} \\ 72s = f {t}^{2} \\ s = \frac{ f{t}^{2} }{72}$

• Hence, value of s is ft²/72.