Home
Class 11
PHYSICS
A car , starting from rest, accelerates ...

A car , starting from rest, accelerates at the rate `f` through a distance `S` then continues at constant speed for time `t` and then decelerates at the rate `(f)/(2)` to come to rest . If the total distance traversed is `15 S` , then

Text Solution

Verified by Experts

Let constant speed be `v_(m)`
for time `t_(1), v_(m)= ft_(1) and S=(1)/(2)ft_(1)^(2)`
for time, `t_(2) O=v_(m) -(f)(2) t_(2)rArr t_(1)=2t_(1)`
`S-(3)=(1)/(2)((f)/(2))t_(2)^(2)=((f)/(4))(4t_(1)^(2)) =ft_(1)^(2)=2S`
Therefore `S+v_(m)t+2S=15SrArrv_(m)t=12SrArr ft_(1)t=12S`
`rArrf((2S)/(f))^(1//2)t=12S rArr 2Sf= (144S^(2))/(t^(2)) rArr S=(ft^(2))/(72)`
Promotional Banner

Topper's Solved these Questions

  • KINEMATICS (MOTION ALONG A STRAIGHT LINE AND MOTION IN A PLANE)

    ALLEN|Exercise BEGINNER S BOX-1|6 Videos

  • KINEMATICS (MOTION ALONG A STRAIGHT LINE AND MOTION IN A PLANE)

    ALLEN|Exercise BEGINNER S BOX-2|7 Videos

  • KINEMATICS

    ALLEN|Exercise Integer Type Question|3 Videos

  • MISCELLANEOUS

    ALLEN|Exercise SUBJECTIVE QUESTION|9 Videos

Similar Questions

Explore conceptually related problems

A car, starting from rest, accelerates at the rate f through a distance s, then continues at constant speed for time t and then decelerates at the rate f/ 2 to come to rest. If the total distance travelled is 15 s, then

A car,starting from rest, accelerates at the rate f through a distance s, then continuous at constant speed for time t and then decelerates at the rate f/2 come to rest. If the total distance traversed is 5 s,then :

A car, starting from rest, accelerates at the rate (f) through a distance (S), then continues at constant speed for some time (t) and then decelerates at the rate f//2 to come to rest. If the total distance is 5 S , then prove that S=1/2/ ft^(@) .

A particle starts from rest, accelerates at 2(m)/(s^2) for 10 s and then goes for constant speed for 30 s and then decelerates at 4(m)/(s^2) . Till it stops. What is the distance travelled by it?

A particle starts from rest accelerates at 2 m//s^2 for 10 s and then goes for constant speed for 30 s and then decelerates at 4 m//s^2 till it stops. What is the distance travelled by it.

A car starting from rest is acceleration at constant rate until it attains a speed of 12 m/s. It moves with the constant speed 12 m/s for some time and then it is retarded at other constant rate until it comes to rest. Considering theta the car moves with constant speed for half of the time of total journey, what will be the average speed of the car for the whole journey (in m/s) ?

A particles starts from rest and has an acceleration of 2m//s^(2) for 10 sec. After that , it travels for 30 sec with constant speed and then undergoes a retardation of 4m//s^(2) and comes back to rest. The total distance covered by the particle is

A car accelerates from rest at a constant rate for some time after which it decelerates at a constant rate beta to come to rest. If the total time elapsed is t, the maximum velocity acquired by the car is given by :