A car starting from rest is
accelerated to a speed of 60m/s in 1min. Find the distance covered by the car in that time interval.
Answers
Answer:
- Distance covered by Car = 1800 m
Explanation:
Given
- Initial velocity of Car, u = 0
- Final velocity of Car, v = 60 m/s
- Time taken for change in velocity, t = 1 min = 60 s
To find
- Distance covered by Car in the given time interval, s =?
Formula required
- First equation of motion
v = u + a t
- Second equation of motion
s = u t + 1/2 a t²
[ Where v is final velocity, u is initial velocity, t is time taken ,a is acceleration, and s is distance covered ]
Solution
Let, acceleration of Car be a
Using first equation of motion
→ v = u + a t
→ 60 = 0 + a × 60
→ 60 = 60 a
→ a = 60/60
→ a = 1 m/s²
Now, Using second equation of motion
→ s = u t + 1/2 a t²
→ s = 0 × 60 + 1/2 × 1 × 60²
→ s = 1/2 × 3600
→ s = 1800 m
Therefore,
- Distance covered by Car in this time interval of 1 min is 1800 metres.
Answer:
:
Given : -
- Velocity v = 60
- Time t = 1 min = 60 sec
To Find : -
- the distance covered by the car in that time interval.
Solution : -
(I). The average velocity is
v_{av} v_1 + v _ f / 2
v_{av} 0 + 16.66/2
v_{av}=8.33
= 8.33 m/s
(2). The acceleration is
Using equation of motion
v = u+at
a = v - u / t
a= 16.66/ 60
a = 0.278 m/s
(3). The distance traveled in first 1 mint
Using the distance formula
d = v × t
d = 16.66 × 60
d = 999.6 m/s
Hence, The average velocity is 8.33 m/s, the acceleration is 0.278 m/s² and the distance traveled in first 1 min is 999.6 m/s