car accelerates from 6 ms–1 16 ms–1 in 10 sec. Calculate(a) the acceleration and(b) the distance covered by the car at that time
Question
car accelerates from 6 ms–1 16 ms–1 in 10 sec. Calculate(a) the acceleration and(b) the distance covered by the car at that time
Solution 1
(a) Acceleration is defined as the rate of change of velocity per unit of time. It can be calculated using the formula:
a = (v_f - v_i) / t
where: v_f = final velocity v_i = initial velocity t = time
Substituting the given values:
a = (16 ms–1 - 6 ms–1) / 10 s
a = 10 ms–2
So, the acceleration of the car is 10 meters per second squared.
(b) The distance covered by the car can be calculated using the formula:
d = v_i*t + 0.5*a*t^2
Substituting the given values:
d = 6 ms–1 * 10 s + 0.5 * 10 ms–2 * (10 s)^2
d = 60 m + 500 m
d = 560 m
So, the car covered a distance of 560 meters.
Solution 2
(a) Acceleration is defined as the rate of change of velocity per unit of time. It can be calculated using the formula:
a = (v_f - v_i) / t
where: v_f = final velocity v_i = initial velocity t = time
Substituting the given values:
a = (16 ms–1 - 6 ms–1) / 10 s
a = 10 ms–2
So, the acceleration of the car is 10 ms–2.
(b) The distance covered by the car can be calculated using the formula:
d = v_i*t + 0.5*a*t^2
Substituting the given values:
d = 6 ms–1 * 10 s + 0.5 * 10 ms–2 * (10 s)^2
d = 60 m + 500 m
d = 560 m
So, the car covered a distance of 560 meters.
Solution 3
(a) Acceleration is defined as the rate of change of velocity per unit of time. It can be calculated using the formula:
a = (v_f - v_i) / t
where: a is acceleration, v_f is the final velocity, v_i is the initial velocity, and t is the time.
Substituting the given values into the formula:
a = (16 ms–1 - 6 ms–1) / 10 s
a = 10 ms–2
So, the acceleration of the car is 10 meters per second squared.
(b) The distance covered by the car can be calculated using the formula:
d = v_i*t + 0.5*a*t^2
where: d is the distance, v_i is the initial velocity, a is acceleration, and t is the time.
Substituting the given values into the formula:
d = 6 ms–1 * 10 s + 0.5 * 10 ms–2 * (10 s)^2
d = 60 m + 500 m
d = 560 m
So, the car covered a distance of 560 meters.
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