Guest

At any instant T = 0 motorbike start from rest in a given resistance a car moving with constant speed overtakes the motorbike at instant when it is moving with a speed 40 m / second motorbike accelerates uniformly till T=18 then moves with constant speed and vortex the car at T = 27 1. Find out the maximum speed of motorbike 2. Find out the acceleration of bike 3. The maximum separation between the car and the bike before overtake the car

At any instant T = 0 motorbike start from rest in a given resistance a car moving with constant speed overtakes the motorbike at instant when it is moving with a speed 40 m / second motorbike accelerates uniformly till T=18 then moves with constant speed and vortex the car at T = 27 
1. Find out the maximum speed of motorbike
2. Find out the acceleration of bike
3. The maximum separation between the car and the bike before overtake the car

Grade:10

1 Answers

Saurabh Koranglekar
askIITians Faculty 10335 Points
3 years ago
distance travelled by car

v= 40m/s

t = 27 sec

distance = 40 × 27

= 1080 m

speed of bike at 18 sec will be

v = 18a

distance covered at this speed with a acceleration in 18 sec will be

v² = u²+ 2 as¹

(18a)² = 2as¹

324a² = 2as¹

s¹ = 162 a

distance covered in 9 sec at the speed of 18a

s² = 9 ×18a

s² = 162 a

now we know the total distance which is 1080 and it will be equal to the sum of both distance we calculated

1080 = 162 a + 162 a

1080 = 324 a

a = 1080/324

a = 3.3333 m/s²

now the max speed of the bike at 18 sec will be

v = 18a

v = 18 × 3.333

v = 59.99 m/s

and the gap between bike and car will be

distance covered by car in 18 sec - 162 a

=18 × 40 - 162 × 3.333

=180 m


Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free