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Byju's Answer

Standard XII

Physics

1st Equation of Motion

A car moving ...

Question

A car moving with constant acceleration covered the distance between two points 60.0 m apart in 6.00 s. Its speed as it passes the second point was 15.0 m/s.What was the speed at the first point?

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Solution

$v = 15 m / s, s = 60 m, t = 6 s v = u + a t \Rightarrow v t = u t + a t^{2} . . . . (1) s = u t + \frac{1}{2} a t^{2} . . . . . (2) v t - s = \frac{1}{2} a t^{2} a = \frac{v t - s}{\frac{1}{2} t^{2}} = \frac{90 - 60}{18} = \frac{5}{3} m / s^{2} u = v - a t = 15 - \frac{5}{3} \times 6 = 5 m / s$

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A car moving with constant acceleration covered the distance between two points 60.0 m apart in 6.00 s. Its speed as it passes the second point was 15.0 m/s.What was the speed at the first point?

v=15m/s,s=60m,t=6sv=u+at⇒vt=ut+at2....(1)s=ut+12at2.....(2)vt−s=12at2a=vt−s12t2=90−6018=53m/s2u=v−at=15−53×6=5m/s

$v = 15 m / s, s = 60 m, t = 6 s v = u + a t \Rightarrow v t = u t + a t^{2} . . . . (1) s = u t + \frac{1}{2} a t^{2} . . . . . (2) v t - s = \frac{1}{2} a t^{2} a = \frac{v t - s}{\frac{1}{2} t^{2}} = \frac{90 - 60}{18} = \frac{5}{3} m / s^{2} u = v - a t = 15 - \frac{5}{3} \times 6 = 5 m / s$