We can use the formula:
retarding force = (mass × change in velocity) ÷ time
We need to first find the time it took for the car to come to rest, which can be found using the equation:
final velocity = initial velocity + (acceleration × time)
Since the final velocity is 0 m/s (it's come to rest), we can rearrange the equation to solve for time:
time = (final velocity - initial velocity) ÷ acceleration
Since the car is braking, the acceleration is negative, so:
time = (0 - 14.0) ÷ (-a)
where a is the acceleration.
We can use the distance formula:
distance = (initial velocity × time) + (0.5 × acceleration × time²)
Since we know the distance (55.0 m) and the initial velocity (14.0 m/s), we can solve for the acceleration:
55.0 = (14.0 × t) + (0.5 × a × t²)
Using the formula for time we found earlier, we can substitute and solve for a:
55.0 = (14.0 × [(0 - 14.0) ÷ (-a)]) + (0.5 × a × [(0 - 14.0) ÷ (-a)]²)
Simplifying:
55.0 = (14.0 × (14.0 ÷ a)) + (0.5 × a × (14.0 ÷ a)²)
Multiplying through by a²:
55.0a² = (14.0 × a × 14.0) + (0.5 × a × 14.0²)
Simplifying further:
55.0a² = 196a + 98a
55.0a² = 294a
Dividing through by a:
55.0a = 294
a = 5.345 m/s²
Now that we have the acceleration, we can use the first formula to find the retarding force:
retarding force = (mass × change in velocity) ÷ time
The change in velocity is just the initial velocity (14.0 m/s), since the car comes to rest.
The time is the time it took for the car to come to rest, which we found earlier:
time = (0 - 14.0) ÷ (-a) = 2.616 s
Substituting:
retarding force = (760 × 14.0) ÷ 2.616
retarding force = 4051.72 N
Therefore, the retarding force on the car is 4051.72 N.
A car of mass 760kg travel with a velocity of 14.0ms-1 . If the brakes are applied , the car travels a distance of 55.0m before coming to rest . Calculate the retarding force on the car .
3 answers
A baby girl of mass 10kg runs at a speed of 2.0ms-1 . If her acceleration is 0.5ms-1 , calculate (i) the force on her
( ii) the time taken by her to cover a distance of 50m .
( ii) the time taken by her to cover a distance of 50m .
(i) The force on the baby girl can be calculated using the formula:
force = mass × acceleration
Substituting the given values:
force = 10 kg × 0.5 ms^-1
force = 5 N
Therefore, the force on the baby girl is 5 N.
(ii) We can use the formula:
distance = (initial velocity × time) + (0.5 × acceleration × time²)
to find the time taken by the baby girl to cover a distance of 50 m.
Since the baby girl is running at a constant acceleration, we can assume the acceleration is the same throughout the 50 m distance.
Substituting the given values:
50 = (2.0 × t) + (0.5 × 0.5 × t²)
Simplifying:
50 = 2t + 0.25t²
Multiplying through by 4:
200 = 8t + t²
Rearranging:
t² + 8t - 200 = 0
We can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b² - 4ac)) ÷ 2a
where a = 1, b = 8, and c = -200
Substituting:
t = (-8 ± √(8² - 4 × 1 × -200)) ÷ 2 × 1
t = (-8 ± √(864)) ÷ 2
Taking only the positive root (since time cannot be negative):
t = (-8 + √864) ÷ 2
t = 7.37 s (rounded to two decimal places)
Therefore, the time taken by the baby girl to cover a distance of 50 m is 7.37 s.
force = mass × acceleration
Substituting the given values:
force = 10 kg × 0.5 ms^-1
force = 5 N
Therefore, the force on the baby girl is 5 N.
(ii) We can use the formula:
distance = (initial velocity × time) + (0.5 × acceleration × time²)
to find the time taken by the baby girl to cover a distance of 50 m.
Since the baby girl is running at a constant acceleration, we can assume the acceleration is the same throughout the 50 m distance.
Substituting the given values:
50 = (2.0 × t) + (0.5 × 0.5 × t²)
Simplifying:
50 = 2t + 0.25t²
Multiplying through by 4:
200 = 8t + t²
Rearranging:
t² + 8t - 200 = 0
We can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b² - 4ac)) ÷ 2a
where a = 1, b = 8, and c = -200
Substituting:
t = (-8 ± √(8² - 4 × 1 × -200)) ÷ 2 × 1
t = (-8 ± √(864)) ÷ 2
Taking only the positive root (since time cannot be negative):
t = (-8 + √864) ÷ 2
t = 7.37 s (rounded to two decimal places)
Therefore, the time taken by the baby girl to cover a distance of 50 m is 7.37 s.