At any instant t=0 a motorbike start from rest in a given direction, a car moving with constant speed overtakes the motorbike at that instant when it is moving with a speed 40 m/s. Motor bike accelerates uniformly till t=18s and then moves with constant speed and overtakes the car at t=27s. The maximum speed and separation between car and bike at t=18 in m/s is
Answers
Explanation:
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
Answer:
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