A car accelerates from rest at a constant rate of 2 m/s2 for some time, after that it retards at a constant rate of 3 m/s2 and come to rest. If the total time elapsed is 8 s, then the maximum velocity acquired by the car is
9.6 m/s
4.8 m/s
2.4 m/s
12 m/s
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Answer:
The maximum velocity acquired by the car is 9.6 m/s.
Explanation:
Given,
[tex]a_{1} = 2 m/s^{2}[/tex]
[tex]a_{2} = 3 m/s^{2}[/tex]
The total time elapsed (t) = 8 s
Let,
v = The maximum velocity attained.
[tex]t_{1}[/tex] = Time for which car accelerated
[tex]t_{2}[/tex] = Time for which car retarded.
∴[tex]t_{1} = \frac{v-u}{a_{1} }[/tex] and [tex]t_{2} = \frac{v-u}{a_{2} }[/tex]
[tex]= \frac{v-0}{2}[/tex] [tex]= \frac{0-v}{-3}[/tex]
[tex]=\frac{v}{2}[/tex] [tex]=\frac{v}{3}[/tex]
Now,
The total time elapsed (t) = 8 s
⇒ [tex]t_{1} + t_{2} = 8[/tex]
⇒ [tex]\frac{v}{2} +\frac{v}{3} =8[/tex]
⇒ [tex]\frac{3v+2v}{6} = 8[/tex]
⇒ 5v = 48
∴ v = 9.6 m/s
The maximum velocity acquired by the car is 9.6 m/s.