A particle is moving in a straight line with initial velocity u and uniform acceleration a. If the sum of the distances travelled in the t th second and (t+1) th seconds is 100 cm, what is its velocity after t seconds in cm/s?
s = u + a(2n - 1)/2That's all you need, believe me.It's the expression that gives the distance travelled by a body in the nth second, if the acceleration is uniform.The velocity of such an object after t seconds is given byv = u + atSo the sum of the distances travelled during the t-th and (t+1)th second amount to 100cmThen:100 = s(t-th) + s(t+1th)100 = u + a(2t - 1)/2 + u + a(2t + 2 - 1)/2100 = 2u + a(2t - 1)/2 + a(2t + 1)/2100 = 2u + (a/2)(2t - 1 + 2t + 1)100 = 2u + (a/2)(4t)100 = 2u + a(2t)100 = 2u + 2at100 = 2(u + at)50 = u + at50 = vTherefore, the velocity after t seconds is 50cm/s.
A train on a straight, level track has an initial speed of 45 km/h. a uniform acceleration of 1.50 m/s^2 is?
<< what is the speed of the train at the end of this distance? >> Your working equation is Vf^2 - Vo^2 = 2as where Vf = final velocity Vo = initial velocity = 45 km/hr. = 12.5 m/sec (given) a = acceleration = 1.5 m/sec^2 s = distance travelled = 200 meters Substituting appropriate values, Vf^2 - 12.5^2 = 2(1.5)(200) Vf^2 = 12.5^2 + 2(1.5)(200) Solving for Vf, Vf = 27.5 m/sec = 99 km/hr. << how long did it take for the trian to travel the 200m? >> There are several options on how to solve for this travel time of the train but the easiest way is to use the formula a = (Vf - Vo)/T where T = time for train to travel 200 meters and all the other terms have been previously defined Substituting values, 1.50 = (27.5 - 12.5)/T Solving for T, T = 15/1.5 T = 10 seconds
A person traveling on a straight line moves with velocity [math]v_1[/math] for a distance x and with a uniform velocity [math]v_2[/math] for the next equal distance. What is the average velocity v given by?
As you know ,v =x/t ,where v=velocity,x=distance travelled , t=time takenHence , t1= x/v1Similarly ,t2=x/v2Therefor, avg. velocity,v = x+x/((x/v1)+(x/v2))v= 2x(v1*v2)/x(v1+v2)v=x(v1*v2)/(v1+v2)
The initial velocity of a body is 7m/s along a straight line. It has a uniform acceleration of 4m/s2. What is the distance covered by the body in the 5th second of its motion?
Using the formula: [math]s(t)=ut+\frac{1}{2}a t^2[/math] gives the distance [math]s[/math] covered up to a specific time [math]t[/math] which is derived in Physics: Where does the half come from in the equation of motion. You know [math]u=7[/math], [math]a=4[/math] and [math]t=5[/math] so you just need to substitute these into the formula. The question asks specifically for the distance that is covered in the 5th second only, that is, ignoring the distance covered prior to this time. This can be worked out by working out the distance covered up to [math]t=5[/math] and subtracting away the distance covered up to [math]t=4[/math] thereby leaving the distance covered in the 5th second. Therefore we need: [math]s(5)-s(4)=[/math][math]7\times 5+\frac{1}{2}\times 4\times 5^2-(7\times 4+\frac{1}{2}\times 4\times 4^2)[/math]which is equal to:[math]85-60=25\; \; \text{meters}[/math]Indeed you can work out the distance covered in any second [math]x[/math] using:[math]s(x)-s(x-1)=[/math][math] ux+\frac{1}{2}ax^2-\left(u (x-1)+\frac{1}{2} a (x-1)^2\right) [/math]EDIT the simplification steps: first distribute the negative over the brackets:[math]ux+\frac{1}{2}ax^2-u (x-1)-\frac{1}{2} a (x-1)^2[/math]then expand out the brackets:[math]ux+\frac{1}{2}ax^2-ux+u-\frac{1}{2} a (x^2-2x+1)[/math]notice that the [math]ux[/math] cancels, then expand out bracket again:[math]\frac{1}{2}ax^2+u-\frac{1}{2} ax^2+ax-\frac{1}{2}a[/math]then after cancelling terms gives:[math]u+ax-\frac{1}{2}a[/math]which you can of course check by using the information from your question and it gives 25 meters too.
Physics help??? Train traveling along a horizontal track?
A science student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of 12.8 m/s. The student throws a ball along a path that she judges to make an initial angle of 30.7° with the horizontal and to be in line with the track. The student's professor, who is standing on the ground nearby, observes the ball to rise vertically. How high does the ball rise? (Neglect air resistance.) I keep getting this guy wrong.. Any help would be greatly appreciated
A car moves with a constant velocity of 10 m/s for 10s along a straight track, then it moves with uniform acceleration of 2 m/s2 for 5s. What are the total displacement and velocity at the end of the 5th second of acceleration?
Here, In first case,S=vt= 10m/s ×10s= 100 mIn second case,Now, here the velocity beocomes initial velocity.By 2nd equ. of motionS = ut+1/2at^2= 10×5 +1/2×2×(5)^2= 50+25= 75mTotal displacement = (displacement in 1st case) + (displacement in 2nd case)= 100m +75m= 175 mNow final velocity,By 1st equ. of motionv=u+at= 10+2×5= 10+10= 20m/s
A body moving along a straight path at a velocity of 20 m/s attains and acceleration of 4 m/s square. What is the velocity of the body after 2 seconds?
Quora IS NOT intended to help you with your homework.Read your Physics book instead !Ok … I cannot resist…the answer should be 28 m/s.:)
A drag racing car starts from rest at [math]t = 0[/math] and moves along a straight line with a velocity given by [math]v = bt^2[/math], where b is a constant. What is the expression for the distance traveled by this car from its position at [math]t=0[/math]?
Since we already know the velocity of the object to be [math]bt^2[/math], we just need to convert this to a distance function. The distance function is just the integral of the velocity function.[math] \int_0^tbt^2\,dt [/math][math]=bt^3/3[/math]
A science student is riding on a flatcar of a train traveling along a straight, horizontal track at a constant?
Because of the observation of the professor, the student must have been facing to the rear. The horizontal component of the velocity, Vh, of the throw must have nullified the train's horizontal speed forward. So Vh = V*cos64 = -12.4 m/s where V is the velocity as observed by the student. Solve for V. The vertical component of the velocity would be Vv = V*sin64 Use the value of Vv as the initial velocity in the kinematic formula Vf^2 = Vi*2 + 2*a*y where Vf = 0 (that's the condition where it has reached as high as it can) a = -g (that's the acceleration in the vertical direction, negative because the initial speed is upward) y is your result.
A science student is riding on a flatcar of a train traveling along a straight horizontal track at a constant?
We're told that the throw is in line with the track, but not whether it's forward or reverse (in the train's direction of travel, or opposite to it). But that should be obvious, because the only way for the toss to be vertical in the ground frame is for it to be toward the rear in the train frame. And so the horizontal component, v[x], of the throw is, in the train frame: v[x] = v cosθ with θ = 38.8º, and v = speed of throw in the train frame. and the horizontal component is 0 in the ground frame, which differs from the train frame by V = 12.2 m/s, so: v[x] = v cosθ = V, v = V/cosθ Then the vertical component, which is what determines the maximum height reached by the ball, is v[z] = v sinθ = V tanθ and the maximum height, H, is found from: gH = ½(v[z])² = ½V²tan²θ H = V²tan²θ /2g = 12.2² • 0.80402² /(2*9.807) m = 4.91 m (Note carefully: This is the height above the ball's release point, NOT the ground!)