A car moving with a speed of 25 m/s takes a U-tum in 5 s, without changing its speed. The average acceleration during these 5 s is

Question

A car moving with a speed of 25 m/s takes a U-tum in 5 s, without changing its speed. The average acceleration during these 5 s is (A)  10 m/s2  (B)  5 m/s2  (C)  2 m/s2  (D)  7 m/s2

Tardigrade Admin · Accepted Answer

Average acceleration text= ( textchange in velocity/ texttime) =(25-(-25)/5) =10 m/s2

Q. A car moving with a speed of 25 m/s takes a U-tum in 5 s, without changing its speed. The average acceleration during these 5 s is

$10 m / s^{2}$

$5 m / s^{2}$

$2 m / s^{2}$

$7 m / s^{2}$

Average acceleration $= \frac{change in velocity}{time}$ $= \frac{25 - ( - 25 )}{5}$ $= 10 m / s^{2}$

Q. A car moving with a speed of 25 m/s takes a U-tum in 5 s, without changing its speed. The average acceleration during these 5 s is

10m/s2

5m/s2

2m/s2

7m/s2

Average acceleration = timechange in velocity​ =525−(−25)​ =10m/s2

$10 m / s^{2}$

$5 m / s^{2}$

$2 m / s^{2}$

$7 m / s^{2}$

Average acceleration $= \frac{change in velocity}{time}$ $= \frac{25 - ( - 25 )}{5}$ $= 10 m / s^{2}$