A car is travelling along the road at 8m/s.IT accelerates at 1m/s for a distance of 18m.how fast is it then travelling?
U=8m/s
a=1m/s2
s=18m
v=?
we know
v2=u2+2as
v2=82+2*1*18
v2=100
v=10m/s
it is travelling at a speed of 10m/s
U=0
A=2m/S2
T=10s
V=u+at
V=0+2×10
U=20m/s
S=ut+1/2 at2
=0×10+1/2×10×10=100m
To find the final speed of the car, we need to calculate the additional distance covered during the acceleration. The initial speed of the car is 8 m/s, and it accelerates at a rate of 1 m/s². The additional distance traveled during this acceleration can be calculated using the equation:
distance = (initial velocity × time) + (0.5 × acceleration × time²)
In this case, we know the initial velocity is 8 m/s, acceleration is 1 m/s², and the distance is 18 m. However, we need to find the time taken for this acceleration.
To find the time, we can rearrange the equation for distance and solve for time:
distance = (initial velocity × time) + (0.5 × acceleration × time²)
18 = (8 × time) + (0.5 × 1 × time²)
18 = 8t + 0.5t²
Rearrange the equation to form a quadratic equation:
0.5t² + 8t - 18 = 0
Next, you can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
In this case, a = 0.5, b = 8, and c = -18. Substituting these values into the formula gives:
t = (-8 ± √(8² - 4(0.5)(-18)) / (2(0.5))
t = (-8 ± √(64 + 36)) / 1
t = (-8 ± √100) / 1
t = (-8 ± 10) / 1
Solving for t gives two solutions:
t₁ = (-8 + 10) / 1 = 2
t₂ = (-8 - 10) / 1 = -18
Since time cannot be negative, we discard the negative solution. Therefore, the time taken for acceleration is 2 seconds.
Now, we can use the final time (2 seconds) to find the final speed of the car. During this time, the car accelerates at a rate of 1 m/s². We can use the equation:
final velocity = initial velocity + (acceleration × time)
Substituting the values, we get:
final velocity = 8 + (1 × 2)
final velocity = 8 + 2
final velocity = 10 m/s
Therefore, the car is traveling at a speed of 10 m/s after accelerating for a distance of 18 m.