A driver of a car travelling along a straight road with a speed of 72k.m./hour . Observes a sign board which gives the speed limit 54k.m./h.the signboard is 70m ahead when the driver applies break . Calculate the acceleration the car which will cause the car to pass the signboard at started?
To calculate the acceleration needed for the car to pass the signboard at the stated speed, we can use the following formula:
\(v^2 = u^2 + 2as\)
Where:
- \(v\) is the final velocity of the car (54 km/h)
- \(u\) is the initial velocity of the car (72 km/h)
- \(a\) is the acceleration
- \(s\) is the distance covered (70 m)
First, convert the velocities from km/h to m/s:
\(72 \, \text{km/h} = 72 \times \frac{1000}{3600} \, \text{m/s} = 20 \, \text{m/s}\)
\(54 \, \text{km/h} = 54 \times \frac{1000}{3600} \, \text{m/s} = 15 \, \text{m/s}\)
Now, substitute the given values into the formula and solve for \(a\):
\(15^2 = 20^2 + 2a \times 70\)
\(225 = 400 + 140a\)
Rearranging the equation:
\(140a = 225 - 400\)
\(140a = -175\)
\(a = -\frac{175}{140}\)
\(a \approx -1.25 \, \text{m/s}^2\)
Therefore, the acceleration needed for the car to pass the signboard at the stated speed is approximately \(-1.25 \, \text{m/s}^2\).
To calculate the acceleration required for the car to pass the signboard when the driver applies the brakes, we need to use the equations of motion.
Let's break down the information given:
Initial speed of the car, u = 72 km/h
Final speed of the car, v = 54 km/h
Distance between the car and the signboard, s = 70 m
Acceleration of the car, a = ?
First, we need to convert the speeds from km/h to m/s:
Initial speed of the car, u = 72 km/h = 72 * (1000/3600) m/s = 20 m/s
Final speed of the car, v = 54 km/h = 54 * (1000/3600) m/s = 15 m/s
Now, let's use the formula:
v^2 = u^2 + 2as
Rearranging the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
Substituting the values:
a = (15^2 - 20^2) / (2 * 70)
a = (225 - 400) / 140
a = -175 / 140
a ≈ -1.25 m/s^2
Therefore, the acceleration required for the car to pass the signboard when the driver applies the brakes is approximately -1.25 m/s^2. The negative sign indicates that the car is decelerating or slowing down.
A car is travelling on a straight road which has a speed limit of 95.0 km/h. The driver can see that just ahead the speed limit changes to 50.0 km/h. If she is currently travelling at the speed limit and she accelerates with a magnitude of 3.00 m/s2, what distance does the car go while slowing down to the new speed limit?
72t + 1/2 at^2 = 70
72 + at = 54
so, a = -16.2 m/s^2 applied for t=10/9 seconds