ac = V ² = 0.7 ms⁻² aₜ = 0.5 ms⁻² a = √(2^2 c+a²)ₜ=0.86⁻² If θ is the angle between the net acceleration and the velocity of the cyclist, then θ = tan⁻¹(ac+ aₜ) = tan(_^(-1)) (1.4) =54°28'
ac = V ² = 0.7 ms⁻²
aₜ = 0.5 ms⁻²
a = √(2^2 c+a²)ₜ=0.86⁻²
If θ is the angle between the net acceleration and the velocity of the cyclist,
then
θ = tan⁻¹(ac+ aₜ) = tan(_^(-1)) (1.4) =54°28′
A cyclist is riding with a speed of 27 kmh⁻¹ . As he approaches a circular turn on the road of radius 30 m, he applies brakes and reduces his speed at the constant rate 0.5 ms⁻²). What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?
ac = V ² = 0.7 ms⁻² aₜ = 0.5 ms⁻² a = √(2^2 c+a²)ₜ=0.86⁻² If θ is the angle between the net acceleration and the velocity of the cyclist, then θ = tan⁻¹(ac+ aₜ) = tan(_^(-1)) (1.4) =54°28'
ac = V ² = 0.7 ms⁻²
See lessaₜ = 0.5 ms⁻²
a = √(2^2 c+a²)ₜ=0.86⁻²
If θ is the angle between the net acceleration and the velocity of the cyclist,
then
θ = tan⁻¹(ac+ aₜ) = tan(_^(-1)) (1.4) =54°28′
A bullet fired at an angle of 30° with the horizontal hits the ground 3 km away. By adjusting the angle of projection, can one hope to hit the target 5 km away ? Assume that the muzzle speed to be fixed and neglect air resistance.
Maximum Range = 3.46 km So it is not possible.
Maximum Range = 3.46 km
See lessSo it is not possible.
A hiker stands on the edge of a clift 490 m above the ground and throws a stone horizontally with an initial speed of 15 ms⁻¹. Neglecting air resistance, find the time taken by the stone to reach the ground and the speed with which it hits the ground. (g = 9.8 ms⁻²)
time = 10 seconds V=√(Vₓ²+Vᵧ²) + √(15²+98²) =99.1 m/s⁻¹
time = 10 seconds
See lessV=√(Vₓ²+Vᵧ²) + √(15²+98²) =99.1 m/s⁻¹
An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft position 10 second apart is 30°, what is the speed of the aircraft ?
Speed = 182.2 ms⁻¹
Speed = 182.2 ms⁻¹
See lessA motorboat is racing towards north at 25 kmh⁻¹ and the water current in that region is 10 kmh⁻¹ in the direction of 60° east of south. Find the resultant velocity of the boat.
V = 21.8 kmh⁻¹ angle with north, θ = 23.4
V = 21.8 kmh⁻¹
See lessangle with north, θ = 23.4