Question

The speed-time graph of a ball moving along a straight line is shown in the figure
below. Calculale the acceleration that brings the ball to rest.
Help me​

Answer 1

Answer:

acceleration of the ball is equal to the area of the triangle

1/2*b*h

1/2*50*4=100m/s^2

Answer 2

Answer:

The speed time graph of a ball of mass 30 g 30g moving along a straight line is shown in (figure). Calculate the opposing force that brings the ball to rest. Read more on Sarthaks.com - https://www.sarthaks.com/1160534/speed-graph-mass-moving-along-straight-line-shown-figure-calculate-opposing-force-brings

Explanation:

Here m = 30 g = 0.03 k g , F = ? m=30g=0.03kg,F=? From the graph, we find u = 50 m / s , v = 0 , t = 4 s u=50m/s,v=0,t=4s As v = u + a t v=u+at ∴ 0 = 50 + a × 4 , a = − 50 4 = − 12.5 m / s 2 ∴0=50+a×4,a=-504=-12.5m/s2 F = m a = 0.03 × ( − 12.5 ) = − 0.37 N F=ma=0.03×(-12.5)=-0.37NRead more on Sarthaks.com - https://www.sarthaks.com/1160534/speed-graph-mass-moving-along-straight-line-shown-figure-calculate-opposing-force-brings

Hope it will help you.