A car starts from rest and attains a velocity of 10m/s in 40s. The driver applies breaks and slows down the car to 5m/s in 10s. Find the acceleration of the car in both the cases.
Answers
Concept
Recall the first equation of motion
v = u +at
where u is initial velocity, a is acceleration, v is final velocity and t is time.
Given
A driver starts the car from rest and attains a velocity of 10m/s in 40s and applies breaks and slows down the car to 5m/s in 10s.
To find
We have to find the acceleration of the car in both cases.
Solution
(i) When the driver starts the car
By using the first equation of motion
v = u+at
10 = 0+a(40)
a = 1/4 = 0.25 m/s²
(ii) When he applies breaks and slows down the car
By using the first equation of motion
v = u+at
5 = 10+a(10)
-5 = a(10)
a = -0.5 m/s²
Hence, the acceleration of the car in both cases is 0.25 m/s²and -0.5 m/s².
Answer:
Concept
Recall the first equation of motion
v = u +at
where u is initial velocity, a is acceleration, v is final velocity and t is time.
Given
A driver starts the car from rest and attains a velocity of 10m/s in 40s and applies breaks and slows down the car to 5m/s in 10s.
To find
We have to find the acceleration of the car in both cases.
Solution
(i) When the driver starts the car
By using the first equation of motion
v = u+at
10 = 0+a(40)
a = 1/4 = 0.25 m/s²
(ii) When he applies breaks and slows down the car
By using the first equation of motion
v = u+at
5 = 10+a(10)
-5 = a(10)
a = -0.5 m/s²
Hence, the acceleration of the car in both cases is 0.25 m/s²and -0.5 m/s².
Explanation:
easy