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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.
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.
A bike starts at rest and after 5s it is moving at 5mk what is the average acceleration?
v = u + at where u = starting velocity, v = final velocity, a = acceleration, t = time. Here u = 0 so v = at ie a = v/t Now, v = 5 m/s (what is mk?) and t = 5 s So a = (5 m/s) / 5 s = 1 m/s2
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
