10. If a sports car at rest accelerates unifomly to a speed of 144km/h in 5s, it
covers a distance of
100m 2). 140m 3) 60m 4) 80m
Answers
Answer:
Given
u=0
v=144kmph=144×
18
5
=40m/s
t=20sec
We know,
v=u+at
40=a×20
a=2m/s
2
Now,
s=ut+
2
1
at
2
s=
2
1
at
2
=
2
1
×2×20
2
=400m
Answer:
Option 1 : 100 m
Explanation:
Given:
- Initial velocity = 0 m/s
- Final velocity = 144 km/hr = 40 m/s
- Time taken = 5s
To Find:
- Distance covered
Concept:
First we have to find the acceleration of the car. Substituting the data in the second or third equation of motion we get the distance covered by the car.
Solution:
Initially we have to find the acceleration of the car
By the first equation of motion,
v = u + at
where v = final velocity
u = initial velocity
a = acceleration
t = time taken
Substitute the data
40 = 0 + a × 5
5a = 40
a = 40/5
a = 8 m/s²
Hence acceleration of the car is 8 m/s²
Now we have to find the distance covered by the car
By the third equation of motion,
v² - u² = 2as
where v = final velocity
u = initial velocity
a = acceleration
s = distance covered
Substituting the data,
40² - 0² = 2 × 8 × s
16s = 1600
s = 1600/16
s = 100 m
Hence the distance covered by the car is 100 m
Therefore option 1 is correct.