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A car enters a freeway with a speed of 6.5 m s and accelerates to sped of 24 m s in 3.5 min how far does the car travel while is accelerating?
Acceleration = (v - u) / t So a = (24 - 6.5) / 3.5*60 a = 1/12 m/s^2 Now to find the displacement S, just use S = (v+u)(v-u) / 2 * a Hope you would do the rest
A car moving 28 m/s skids to avoid an accident, stopping in 1.8 s. calculate the acceleration of the car while it stops?
u=28m/s v=0m/s t=1.8s a=? Solution: v=u+at 0=28+(-a)1.8 -28=-1.8a 28=1.8a a=15.5m/s2
What is the acceleration of the car between 20 and 50 sec?
the 4 fundamental kinematic equations of motion for constant acceleration (suggest you commit these to memory): s = ut + ½at^2 …. (1) v^2 = u^2 + 2as …. (2) v = u + at …. (3) s = (u + v)t/2 …. (4) where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time. In this case, we know u = 10m/s, v = 50m/s, t = 10s and we want to find a, so we use equation (3) v = u + at 50 = 10 + 10t, so t = (50 – 10)/10 = 4m/s^2
Are you able to divide 3 fractions instead of 2?
Yes. a/b / u/v / r/s = (a*v)/(b*u) / r/s = (a * v * s) / (b * u * r)
A race car accelerates from rest at a rate of 6 m s How fast will the car be going after 5 seconds?
The car's speed after 5 seconds can be calculated using the formula ( v = u + at ), where (v) is the final velocity, (u) is the initial velocity (0 m/s in this case), (a) is the acceleration (6 m/s(^2)), and (t) is the time (5 seconds). Plugging in the values gives (v = 0 + 6 \times 5 = 30 ) m/s. Thus, the car will be going 30 m/s after 5 seconds.
What deceleration is needed to bring a car with a velocity of ms-1 to rest in 80 meters?
The deceleration needed to bring a car to rest can be calculated using the equation: ( v^2 = u^2 + 2as), where (v) is the final velocity (0 m/s), (u) is the initial velocity (given as unknown), (a) is the deceleration, and (s) is the displacement (80 meters). Solving for (a), the deceleration needed would be about 2.52 m/s^2.
Third equation of motion?
v2- u 2 = 2assince, S (Distance) = Average speed x TimeS = U+V / 2 * TS = U+V / 2 * V - U / A {since T = V -U / A}S = V2 - U2 / 2A2AS = V2 - U2OR V2 - U2 = AsHence, Derived.
What is the spelling of survive?
It is spelled s-u-r-v-i-v-e.
Equation for acceleration?
acceleration(a) = (final velocity(v) - Initial velocity(u)) / time (s) Algebraically a = (v - u) / t Where 'v' & 'u' are measured in metres per second ( m/s) or ms^-1 And 't' is the time in seconds measured is 's' Hence a(ms^-2) = v(m/s) - u(m/s)) / t(s) And example is a car starting from rest up to 44 m/s ( 30 mph) in 10 seconds. a = (44 - 0 ) / 10 a = 44/10 a = 4.4 ms^-2. NB Earth's gravitational acceleration(g) is approximately 10 ms^-2.
If you find skid marks 180 meters long and assuming the deceleration was 10 minutes how fast was the car traveling before it hit the brakes?
To determine the initial speed of the car before it hit the brakes, we can use the formula for deceleration: ( v^2 = u^2 + 2as ), where ( v ) is the final velocity (0 m/s when the car stops), ( u ) is the initial velocity, ( a ) is the deceleration, and ( s ) is the distance (180 meters). Assuming a deceleration of 10 m/s² (not minutes), rearranging the formula gives ( u = \sqrt{v^2 - 2as} ). Plugging in the values, we find ( u = \sqrt{0 - 2(-10)(180)} = \sqrt{3600} = 60 ) m/s, which is approximately 216 km/h.
How do you use the LCM to write fractions with a common denominator?
Suppose you have the fractions p/q and r/s. Let the LCM of q and s be t.Then t is a multiple of q as well as of s so let t= q*u and t = s*v Then p/q = (p*u)/(q*u) = (p*u)/t and r/s = (r*v)/(s*v) = (r*v)/t have the same denominators.
If a racing car's speed is 96 ms in 1 s what is the racing car's acceleration?
Assuming the car begins at rest (0 m/s), and it has to reach 96 m/s in 1s, then, by the suvat equations:s = - (displacement)u = 0 ms-1 (initial velocity)v = 96 ms-1 (final velocity)a = ? (acceleration)t = 1s (time)a = (v-u)/ta = (96-0)/1 ms-2a = 96 ms-2
